Let $Q$ denote a queue containing sixteen numbers and $S$ be an empty stack. $\text{Head}(Q)$ returns the element at the head of the queue $Q$ without removing it from $Q$. Similarly $\text{Top}(S)$ returns the element at the top of $S$ without removing it from $S$. Consider the algorithm given below.while Q is not Empty do if S is Empty OR Top(S) <= Head(Q) then x := Dequeue(Q); Push(S, x); else x := Pop(S); Enqueue(Q, x); end endThe maximum possible number of iterations of the while loop in the algorithm is _______.

Let $n$ be the number of elements initially in queue $Q$ (here $n = 16$).

The algorithm sorts the elements into stack $S$ in non-decreasing order from bottom to top (or strictly descending when popped).

Let $I(n)$ denote the maximum number of iterations of the while loop for $n$ elements.

Consider the worst-case scenario, which occurs when the queue elements are in strictly decreasing order (e.g., $n, n-1, \dots, 2, 1$):

1. For the first element, the stack is empty $\rightarrow 1$ Push operation ($1$ iteration).

2. For the $k$-th element ($k = 2, 3, \dots, n$), all $k-1$ elements currently in the stack are strictly greater than the new element. Thus, they must all be popped and enqueued back to $Q$ ($k-1$ Pop iterations), then the new element is pushed ($1$ Push iteration), and finally all $k-1$ popped elements will later be pushed back into the stack ($k-1$ Push iterations).

Thus, processing the $k$-th element adds $(k-1) + 1 + (k-1) = 2k - 1$ iterations.

Summing over all $n$ elements from $k = 1$ to $n$:

Given $n = 16$:

Therefore, the maximum possible number of iterations is 256.