Give you n linklists, each with m numbers inside. the chains index are numbered 0 to (n-1), and the nodes on the initial i-th chain are numbered (i*m) , (i*m+1) ...... (i*m+(m-1))

After that you are given k operations, each operation will select two node numbers a,b. Disconnect the right side of a and the left side of b on the current linklist into four parts, then join a and b together and join the remaining two disconnected parts together.

For example, n=3,m=5. If you choose a,b=1,7 then the result after reconnection is [5,6,2,3,4] and [0,1,7,8,9]

After these operations, ask for all the nodes of the linklist where node x is located and output them from left to right.

First line 4 numbers : n, m, k, x  (n*m<1e6 , 0<k<2e5)

Then follow k lines, each line two numbers: a, b ( 0<=a,b <n*m) meaning the nodes selected

It is guaranteed that the two nodes selected each time are not currently in the same chain

Bonus lab