What circuit or operation corresponds to the tensor product?











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What Clifford gate circuit operating on states $|psi_1rangle$ and $|psi_2rangle$ prepares the state $|Psirangle=|psi_1rangle otimes |psi_2rangle$ ?










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    What Clifford gate circuit operating on states $|psi_1rangle$ and $|psi_2rangle$ prepares the state $|Psirangle=|psi_1rangle otimes |psi_2rangle$ ?










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      What Clifford gate circuit operating on states $|psi_1rangle$ and $|psi_2rangle$ prepares the state $|Psirangle=|psi_1rangle otimes |psi_2rangle$ ?










      share|improve this question















      What Clifford gate circuit operating on states $|psi_1rangle$ and $|psi_2rangle$ prepares the state $|Psirangle=|psi_1rangle otimes |psi_2rangle$ ?







      quantum-gate circuit-model tensor-product






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      edited yesterday









      Blue

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      asked yesterday









      Malcolm Regan

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          I think that you misunderstood the concept of tensor product here. There is no need of a Clifford gate in order to have a multi-qubit system. The fact that a multi-qubit system is described by the tensor produt of their state vectors comes from the fourth postulate of quantum mechanics, refer to the $94^{th}$ page of Nielsen and Chuang to see the formulation of this postulate.



          Consequently, what I mean here is that if you have two qubits represented by state vectors $|psi_1rangle$ and $|psi_2rangle$, the tensor product is the mathematical operation that describes the composite state of the 2-qubit sytem $|Psirangle$. This something known by postulates, and so there is no need to use any Clifford gate to obtain that.



          Clifford gates for example could be used if you have state $|Psirangle=|psi_1rangleotimes |psi_2rangle$ and you want to obtain another state $|Psi'rangle$, then the transformation would be given by some Clifford group unitary so that $|Psi'rangle=U|Psirangle$. However, in order to calculate such unitary, the initial states should be needed, and so here no more can be said.






          share|improve this answer




























            up vote
            5
            down vote













            There is no circuit operation or Clifford gate! If you have $|psi_1rangle$ and $|psi_2rangle$ entering a circuit, then mathematically we say that:
            $$
            |psi_1rangle otimes |psi_2rangle
            $$



            is entering the circuit.



            This means if you represent the whole circuit as a unitary matrix $U$, it will have a dimension corresponding to the size of $|psi_1rangle otimes |psi_2rangle$, not of $|psi_1rangle$ or $|psi_2rangle$. Your output is literally the matrix-vector product:



            $$
            Uleft(left|psi_1rightrangle otimes |psi_2rangleright),
            $$



            even without using any extra Clifford gates!






            share|improve this answer










            New contributor




            G Fleming is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
            Check out our Code of Conduct.

























              up vote
              2
              down vote













              The tensor product is not a gate, but rather a way for us as humans to model the behavior of a quantum system. Whenever we're using multiple qbits, we can look at them in two ways: in their product state (a complex vector of size $2^n$ for $n$ qbits) or their individual state ($|psi_0rangle otimes ldots otimes |psi_{n-1}rangle$). We can usually switch back and forth between the two representations at will (again, they are two equivalent ways of writing the same quantum state) except for when the qbits become entangled, in which case we cannot factor them back into the individual state (this is roughly the definition of entanglement).



              When you're just learning quantum computing and are writing out matrices & vectors explicitly, you usually use the product state to calculate the action of an operator on the quantum state, then factor the result back into the individual state to see the action of that operator on the individual qbits. So for example, if we want to calculate the action of CNOT on $|10rangle$ with leftmost qbit as control:



              $C|10rangle = C begin{bmatrix} 0 \ 1 end{bmatrix} otimes begin{bmatrix} 1 \ 0 end{bmatrix} =
              begin{bmatrix}
              1 & 0 & 0 & 0 \
              0 & 1 & 0 & 0 \
              0 & 0 & 0 & 1 \
              0 & 0 & 1 & 0 \
              end{bmatrix}
              begin{bmatrix} 0 \ 0 \ 1 \ 0 end{bmatrix} =
              begin{bmatrix} 0 \ 0 \ 0 \ 1 end{bmatrix} =
              begin{bmatrix} 0 \ 1 end{bmatrix} otimes begin{bmatrix} 0 \ 1 end{bmatrix}
              = |11rangle$



              So we see that CNOT on $|10rangle$ flipped the rightmost qbit to $|11rangle$ as expected, because the control (leftmost) qbit was $1$.



              Here's an entangled product state which cannot be factored into its individual state:



              $C_{1,0}H_1|00rangle = begin{bmatrix} frac{1}{sqrt{2}} \ 0 \ 0 \ frac{1}{sqrt{2}} end{bmatrix}$



              If you try to factor that out into the tensor product of two qbits, you will see you cannot.






              share|improve this answer




























                up vote
                0
                down vote













                To put another way, the circuit is just



                enter image description here



                As everyone else said, the tensor product is just the way of constructing a composite quantum system from smaller subsystems, and doesn't require gates as such. But ordering is important, $|psi_1rangleotimes|psi_2rangleneq |psi_2rangleotimes|psi_1rangle$, and there is a standard correspondence in quantum circuits between the first item in the tensor product being on the top wire.






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                  I think that you misunderstood the concept of tensor product here. There is no need of a Clifford gate in order to have a multi-qubit system. The fact that a multi-qubit system is described by the tensor produt of their state vectors comes from the fourth postulate of quantum mechanics, refer to the $94^{th}$ page of Nielsen and Chuang to see the formulation of this postulate.



                  Consequently, what I mean here is that if you have two qubits represented by state vectors $|psi_1rangle$ and $|psi_2rangle$, the tensor product is the mathematical operation that describes the composite state of the 2-qubit sytem $|Psirangle$. This something known by postulates, and so there is no need to use any Clifford gate to obtain that.



                  Clifford gates for example could be used if you have state $|Psirangle=|psi_1rangleotimes |psi_2rangle$ and you want to obtain another state $|Psi'rangle$, then the transformation would be given by some Clifford group unitary so that $|Psi'rangle=U|Psirangle$. However, in order to calculate such unitary, the initial states should be needed, and so here no more can be said.






                  share|improve this answer

























                    up vote
                    6
                    down vote



                    accepted










                    I think that you misunderstood the concept of tensor product here. There is no need of a Clifford gate in order to have a multi-qubit system. The fact that a multi-qubit system is described by the tensor produt of their state vectors comes from the fourth postulate of quantum mechanics, refer to the $94^{th}$ page of Nielsen and Chuang to see the formulation of this postulate.



                    Consequently, what I mean here is that if you have two qubits represented by state vectors $|psi_1rangle$ and $|psi_2rangle$, the tensor product is the mathematical operation that describes the composite state of the 2-qubit sytem $|Psirangle$. This something known by postulates, and so there is no need to use any Clifford gate to obtain that.



                    Clifford gates for example could be used if you have state $|Psirangle=|psi_1rangleotimes |psi_2rangle$ and you want to obtain another state $|Psi'rangle$, then the transformation would be given by some Clifford group unitary so that $|Psi'rangle=U|Psirangle$. However, in order to calculate such unitary, the initial states should be needed, and so here no more can be said.






                    share|improve this answer























                      up vote
                      6
                      down vote



                      accepted







                      up vote
                      6
                      down vote



                      accepted






                      I think that you misunderstood the concept of tensor product here. There is no need of a Clifford gate in order to have a multi-qubit system. The fact that a multi-qubit system is described by the tensor produt of their state vectors comes from the fourth postulate of quantum mechanics, refer to the $94^{th}$ page of Nielsen and Chuang to see the formulation of this postulate.



                      Consequently, what I mean here is that if you have two qubits represented by state vectors $|psi_1rangle$ and $|psi_2rangle$, the tensor product is the mathematical operation that describes the composite state of the 2-qubit sytem $|Psirangle$. This something known by postulates, and so there is no need to use any Clifford gate to obtain that.



                      Clifford gates for example could be used if you have state $|Psirangle=|psi_1rangleotimes |psi_2rangle$ and you want to obtain another state $|Psi'rangle$, then the transformation would be given by some Clifford group unitary so that $|Psi'rangle=U|Psirangle$. However, in order to calculate such unitary, the initial states should be needed, and so here no more can be said.






                      share|improve this answer












                      I think that you misunderstood the concept of tensor product here. There is no need of a Clifford gate in order to have a multi-qubit system. The fact that a multi-qubit system is described by the tensor produt of their state vectors comes from the fourth postulate of quantum mechanics, refer to the $94^{th}$ page of Nielsen and Chuang to see the formulation of this postulate.



                      Consequently, what I mean here is that if you have two qubits represented by state vectors $|psi_1rangle$ and $|psi_2rangle$, the tensor product is the mathematical operation that describes the composite state of the 2-qubit sytem $|Psirangle$. This something known by postulates, and so there is no need to use any Clifford gate to obtain that.



                      Clifford gates for example could be used if you have state $|Psirangle=|psi_1rangleotimes |psi_2rangle$ and you want to obtain another state $|Psi'rangle$, then the transformation would be given by some Clifford group unitary so that $|Psi'rangle=U|Psirangle$. However, in order to calculate such unitary, the initial states should be needed, and so here no more can be said.







                      share|improve this answer












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                      share|improve this answer










                      answered yesterday









                      Josu Etxezarreta Martinez

                      1,226217




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                          up vote
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                          down vote













                          There is no circuit operation or Clifford gate! If you have $|psi_1rangle$ and $|psi_2rangle$ entering a circuit, then mathematically we say that:
                          $$
                          |psi_1rangle otimes |psi_2rangle
                          $$



                          is entering the circuit.



                          This means if you represent the whole circuit as a unitary matrix $U$, it will have a dimension corresponding to the size of $|psi_1rangle otimes |psi_2rangle$, not of $|psi_1rangle$ or $|psi_2rangle$. Your output is literally the matrix-vector product:



                          $$
                          Uleft(left|psi_1rightrangle otimes |psi_2rangleright),
                          $$



                          even without using any extra Clifford gates!






                          share|improve this answer










                          New contributor




                          G Fleming is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                          Check out our Code of Conduct.






















                            up vote
                            5
                            down vote













                            There is no circuit operation or Clifford gate! If you have $|psi_1rangle$ and $|psi_2rangle$ entering a circuit, then mathematically we say that:
                            $$
                            |psi_1rangle otimes |psi_2rangle
                            $$



                            is entering the circuit.



                            This means if you represent the whole circuit as a unitary matrix $U$, it will have a dimension corresponding to the size of $|psi_1rangle otimes |psi_2rangle$, not of $|psi_1rangle$ or $|psi_2rangle$. Your output is literally the matrix-vector product:



                            $$
                            Uleft(left|psi_1rightrangle otimes |psi_2rangleright),
                            $$



                            even without using any extra Clifford gates!






                            share|improve this answer










                            New contributor




                            G Fleming is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                            Check out our Code of Conduct.




















                              up vote
                              5
                              down vote










                              up vote
                              5
                              down vote









                              There is no circuit operation or Clifford gate! If you have $|psi_1rangle$ and $|psi_2rangle$ entering a circuit, then mathematically we say that:
                              $$
                              |psi_1rangle otimes |psi_2rangle
                              $$



                              is entering the circuit.



                              This means if you represent the whole circuit as a unitary matrix $U$, it will have a dimension corresponding to the size of $|psi_1rangle otimes |psi_2rangle$, not of $|psi_1rangle$ or $|psi_2rangle$. Your output is literally the matrix-vector product:



                              $$
                              Uleft(left|psi_1rightrangle otimes |psi_2rangleright),
                              $$



                              even without using any extra Clifford gates!






                              share|improve this answer










                              New contributor




                              G Fleming is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                              Check out our Code of Conduct.









                              There is no circuit operation or Clifford gate! If you have $|psi_1rangle$ and $|psi_2rangle$ entering a circuit, then mathematically we say that:
                              $$
                              |psi_1rangle otimes |psi_2rangle
                              $$



                              is entering the circuit.



                              This means if you represent the whole circuit as a unitary matrix $U$, it will have a dimension corresponding to the size of $|psi_1rangle otimes |psi_2rangle$, not of $|psi_1rangle$ or $|psi_2rangle$. Your output is literally the matrix-vector product:



                              $$
                              Uleft(left|psi_1rightrangle otimes |psi_2rangleright),
                              $$



                              even without using any extra Clifford gates!







                              share|improve this answer










                              New contributor




                              G Fleming is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                              Check out our Code of Conduct.









                              share|improve this answer



                              share|improve this answer








                              edited yesterday









                              Mithrandir24601

                              2,1661831




                              2,1661831






                              New contributor




                              G Fleming is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                              Check out our Code of Conduct.









                              answered yesterday









                              G Fleming

                              714




                              714




                              New contributor




                              G Fleming is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                              Check out our Code of Conduct.





                              New contributor





                              G Fleming is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                              Check out our Code of Conduct.






                              G Fleming is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                              Check out our Code of Conduct.






















                                  up vote
                                  2
                                  down vote













                                  The tensor product is not a gate, but rather a way for us as humans to model the behavior of a quantum system. Whenever we're using multiple qbits, we can look at them in two ways: in their product state (a complex vector of size $2^n$ for $n$ qbits) or their individual state ($|psi_0rangle otimes ldots otimes |psi_{n-1}rangle$). We can usually switch back and forth between the two representations at will (again, they are two equivalent ways of writing the same quantum state) except for when the qbits become entangled, in which case we cannot factor them back into the individual state (this is roughly the definition of entanglement).



                                  When you're just learning quantum computing and are writing out matrices & vectors explicitly, you usually use the product state to calculate the action of an operator on the quantum state, then factor the result back into the individual state to see the action of that operator on the individual qbits. So for example, if we want to calculate the action of CNOT on $|10rangle$ with leftmost qbit as control:



                                  $C|10rangle = C begin{bmatrix} 0 \ 1 end{bmatrix} otimes begin{bmatrix} 1 \ 0 end{bmatrix} =
                                  begin{bmatrix}
                                  1 & 0 & 0 & 0 \
                                  0 & 1 & 0 & 0 \
                                  0 & 0 & 0 & 1 \
                                  0 & 0 & 1 & 0 \
                                  end{bmatrix}
                                  begin{bmatrix} 0 \ 0 \ 1 \ 0 end{bmatrix} =
                                  begin{bmatrix} 0 \ 0 \ 0 \ 1 end{bmatrix} =
                                  begin{bmatrix} 0 \ 1 end{bmatrix} otimes begin{bmatrix} 0 \ 1 end{bmatrix}
                                  = |11rangle$



                                  So we see that CNOT on $|10rangle$ flipped the rightmost qbit to $|11rangle$ as expected, because the control (leftmost) qbit was $1$.



                                  Here's an entangled product state which cannot be factored into its individual state:



                                  $C_{1,0}H_1|00rangle = begin{bmatrix} frac{1}{sqrt{2}} \ 0 \ 0 \ frac{1}{sqrt{2}} end{bmatrix}$



                                  If you try to factor that out into the tensor product of two qbits, you will see you cannot.






                                  share|improve this answer

























                                    up vote
                                    2
                                    down vote













                                    The tensor product is not a gate, but rather a way for us as humans to model the behavior of a quantum system. Whenever we're using multiple qbits, we can look at them in two ways: in their product state (a complex vector of size $2^n$ for $n$ qbits) or their individual state ($|psi_0rangle otimes ldots otimes |psi_{n-1}rangle$). We can usually switch back and forth between the two representations at will (again, they are two equivalent ways of writing the same quantum state) except for when the qbits become entangled, in which case we cannot factor them back into the individual state (this is roughly the definition of entanglement).



                                    When you're just learning quantum computing and are writing out matrices & vectors explicitly, you usually use the product state to calculate the action of an operator on the quantum state, then factor the result back into the individual state to see the action of that operator on the individual qbits. So for example, if we want to calculate the action of CNOT on $|10rangle$ with leftmost qbit as control:



                                    $C|10rangle = C begin{bmatrix} 0 \ 1 end{bmatrix} otimes begin{bmatrix} 1 \ 0 end{bmatrix} =
                                    begin{bmatrix}
                                    1 & 0 & 0 & 0 \
                                    0 & 1 & 0 & 0 \
                                    0 & 0 & 0 & 1 \
                                    0 & 0 & 1 & 0 \
                                    end{bmatrix}
                                    begin{bmatrix} 0 \ 0 \ 1 \ 0 end{bmatrix} =
                                    begin{bmatrix} 0 \ 0 \ 0 \ 1 end{bmatrix} =
                                    begin{bmatrix} 0 \ 1 end{bmatrix} otimes begin{bmatrix} 0 \ 1 end{bmatrix}
                                    = |11rangle$



                                    So we see that CNOT on $|10rangle$ flipped the rightmost qbit to $|11rangle$ as expected, because the control (leftmost) qbit was $1$.



                                    Here's an entangled product state which cannot be factored into its individual state:



                                    $C_{1,0}H_1|00rangle = begin{bmatrix} frac{1}{sqrt{2}} \ 0 \ 0 \ frac{1}{sqrt{2}} end{bmatrix}$



                                    If you try to factor that out into the tensor product of two qbits, you will see you cannot.






                                    share|improve this answer























                                      up vote
                                      2
                                      down vote










                                      up vote
                                      2
                                      down vote









                                      The tensor product is not a gate, but rather a way for us as humans to model the behavior of a quantum system. Whenever we're using multiple qbits, we can look at them in two ways: in their product state (a complex vector of size $2^n$ for $n$ qbits) or their individual state ($|psi_0rangle otimes ldots otimes |psi_{n-1}rangle$). We can usually switch back and forth between the two representations at will (again, they are two equivalent ways of writing the same quantum state) except for when the qbits become entangled, in which case we cannot factor them back into the individual state (this is roughly the definition of entanglement).



                                      When you're just learning quantum computing and are writing out matrices & vectors explicitly, you usually use the product state to calculate the action of an operator on the quantum state, then factor the result back into the individual state to see the action of that operator on the individual qbits. So for example, if we want to calculate the action of CNOT on $|10rangle$ with leftmost qbit as control:



                                      $C|10rangle = C begin{bmatrix} 0 \ 1 end{bmatrix} otimes begin{bmatrix} 1 \ 0 end{bmatrix} =
                                      begin{bmatrix}
                                      1 & 0 & 0 & 0 \
                                      0 & 1 & 0 & 0 \
                                      0 & 0 & 0 & 1 \
                                      0 & 0 & 1 & 0 \
                                      end{bmatrix}
                                      begin{bmatrix} 0 \ 0 \ 1 \ 0 end{bmatrix} =
                                      begin{bmatrix} 0 \ 0 \ 0 \ 1 end{bmatrix} =
                                      begin{bmatrix} 0 \ 1 end{bmatrix} otimes begin{bmatrix} 0 \ 1 end{bmatrix}
                                      = |11rangle$



                                      So we see that CNOT on $|10rangle$ flipped the rightmost qbit to $|11rangle$ as expected, because the control (leftmost) qbit was $1$.



                                      Here's an entangled product state which cannot be factored into its individual state:



                                      $C_{1,0}H_1|00rangle = begin{bmatrix} frac{1}{sqrt{2}} \ 0 \ 0 \ frac{1}{sqrt{2}} end{bmatrix}$



                                      If you try to factor that out into the tensor product of two qbits, you will see you cannot.






                                      share|improve this answer












                                      The tensor product is not a gate, but rather a way for us as humans to model the behavior of a quantum system. Whenever we're using multiple qbits, we can look at them in two ways: in their product state (a complex vector of size $2^n$ for $n$ qbits) or their individual state ($|psi_0rangle otimes ldots otimes |psi_{n-1}rangle$). We can usually switch back and forth between the two representations at will (again, they are two equivalent ways of writing the same quantum state) except for when the qbits become entangled, in which case we cannot factor them back into the individual state (this is roughly the definition of entanglement).



                                      When you're just learning quantum computing and are writing out matrices & vectors explicitly, you usually use the product state to calculate the action of an operator on the quantum state, then factor the result back into the individual state to see the action of that operator on the individual qbits. So for example, if we want to calculate the action of CNOT on $|10rangle$ with leftmost qbit as control:



                                      $C|10rangle = C begin{bmatrix} 0 \ 1 end{bmatrix} otimes begin{bmatrix} 1 \ 0 end{bmatrix} =
                                      begin{bmatrix}
                                      1 & 0 & 0 & 0 \
                                      0 & 1 & 0 & 0 \
                                      0 & 0 & 0 & 1 \
                                      0 & 0 & 1 & 0 \
                                      end{bmatrix}
                                      begin{bmatrix} 0 \ 0 \ 1 \ 0 end{bmatrix} =
                                      begin{bmatrix} 0 \ 0 \ 0 \ 1 end{bmatrix} =
                                      begin{bmatrix} 0 \ 1 end{bmatrix} otimes begin{bmatrix} 0 \ 1 end{bmatrix}
                                      = |11rangle$



                                      So we see that CNOT on $|10rangle$ flipped the rightmost qbit to $|11rangle$ as expected, because the control (leftmost) qbit was $1$.



                                      Here's an entangled product state which cannot be factored into its individual state:



                                      $C_{1,0}H_1|00rangle = begin{bmatrix} frac{1}{sqrt{2}} \ 0 \ 0 \ frac{1}{sqrt{2}} end{bmatrix}$



                                      If you try to factor that out into the tensor product of two qbits, you will see you cannot.







                                      share|improve this answer












                                      share|improve this answer



                                      share|improve this answer










                                      answered yesterday









                                      ahelwer

                                      1,003111




                                      1,003111






















                                          up vote
                                          0
                                          down vote













                                          To put another way, the circuit is just



                                          enter image description here



                                          As everyone else said, the tensor product is just the way of constructing a composite quantum system from smaller subsystems, and doesn't require gates as such. But ordering is important, $|psi_1rangleotimes|psi_2rangleneq |psi_2rangleotimes|psi_1rangle$, and there is a standard correspondence in quantum circuits between the first item in the tensor product being on the top wire.






                                          share|improve this answer

























                                            up vote
                                            0
                                            down vote













                                            To put another way, the circuit is just



                                            enter image description here



                                            As everyone else said, the tensor product is just the way of constructing a composite quantum system from smaller subsystems, and doesn't require gates as such. But ordering is important, $|psi_1rangleotimes|psi_2rangleneq |psi_2rangleotimes|psi_1rangle$, and there is a standard correspondence in quantum circuits between the first item in the tensor product being on the top wire.






                                            share|improve this answer























                                              up vote
                                              0
                                              down vote










                                              up vote
                                              0
                                              down vote









                                              To put another way, the circuit is just



                                              enter image description here



                                              As everyone else said, the tensor product is just the way of constructing a composite quantum system from smaller subsystems, and doesn't require gates as such. But ordering is important, $|psi_1rangleotimes|psi_2rangleneq |psi_2rangleotimes|psi_1rangle$, and there is a standard correspondence in quantum circuits between the first item in the tensor product being on the top wire.






                                              share|improve this answer












                                              To put another way, the circuit is just



                                              enter image description here



                                              As everyone else said, the tensor product is just the way of constructing a composite quantum system from smaller subsystems, and doesn't require gates as such. But ordering is important, $|psi_1rangleotimes|psi_2rangleneq |psi_2rangleotimes|psi_1rangle$, and there is a standard correspondence in quantum circuits between the first item in the tensor product being on the top wire.







                                              share|improve this answer












                                              share|improve this answer



                                              share|improve this answer










                                              answered yesterday









                                              DaftWullie

                                              10.5k1534




                                              10.5k1534






























                                                   

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