In this problem we have two stacks A and B $($recall that a stack allows us to push elements onto it and to pop elements off of it in LIFO order$).$ In what follows, we will use n to denoate the number of elements in stack A and use m to denote the number of elements in stack B$.$ Suppose that we use these stacks to implement the following operations:$\backslash \backslash$

$\backslash$begin$\{$itemize$\}$

$\backslash$item PushA$($x$)$: Push element x onto A$.$

$\backslash$item PushB$($x$)$: Push element x onto B$.$

$\backslash$item BigPopA$($k$)$: Pop min $($n$,$ k$)$ elements from A$.$

$\backslash$item BigPopB$($k$)$: Pop min $($m$,$ k$)$ elements from B$.$

$\backslash$item Move$($k$)$: Repeatedly pop one element from A and push it into B$,$ until either k elements have been moved or A is empty.

$\backslash$end$\{$itemize$\}$

We are using the stacks A and B as black boxes$-$ you may assume that PushA, PushB, BigPopA$(1),$ and BigPopB$(1)$ each take one unit of time $($i$.$e$.,$ it takes one time to step to push or pop a single element$).\backslash \backslash$

Prove that the amortized running time of each operation is O$(1).$ You may use either a potential function $$\backslash$Phi$ $($n$,$ m$)$ or a banking scheme, but must use one or the other.$\backslash \backslash$