In one move, Milica can replace the whole string with $$$\texttt{AA} \ldots \texttt{A}$$$. In her second move, she can replace a prefix of length $$$k$$$ with $$$\texttt{BB} \ldots \texttt{B}$$$. The process takes no more than $$$2$$$ operations. The question remains — when can we do better?

In the hint section, we showed that the minimum number of operations is $$$0$$$, $$$1$$$, or $$$2$$$. We have $$$3$$$ cases:

1. No operations are needed if $$$s$$$ already contains $$$k$$$ characters $$$\texttt{B}$$$.

2. Else, we can use brute force to check if changing some prefix leads to $$$s$$$ having $$$k$$$ $$$\texttt{B}$$$s. If we find such a prefix, we print it as the answer, and use only one operation. There are $$$O(n)$$$ possibilities. implementing them takes $$$O(n)$$$ or $$$O(n^2)$$$ time.

3. Else, we use the two operations described in the hint section.

A fun fact is that only one operation is enough. Can you prove it?

If at some point Milena divides $$$a_i$$$ on $$$x$$$ and $$$y$$$, that will not affect the subarray right from $$$a_i$$$ (by dividing some element, it can only get smaller). That means we can try greedy from right to left.

As we described in the hints, we will sort the array greedily from right to left. Let's assume we have sorted subarray $$$[a_{i+1}, \dots, a_n]$$$, and now we are operating on element $$$a_i$$$. Let $$$a_i$$$ be splited into integers $$$x_1,\dots,x_{m+1}$$$ after $$$m$$$ operations. As we want to make all $$$x_i$$$ as large as possible, to spend as few operations as possible on the left subarray, all $$$x_i$$$ will have the value $$$\lfloor \frac{a_i}{m+1} \rfloor$$$ or $$$\lceil \frac{a_i}{m+1} \rceil$$$. So in order to sort the array, $$$\lfloor \frac{a_i}{m+1} \rfloor \leq a_{i+1}$$$ must hold. Therefore, the minimum value of $$$m$$$ for which it is possible to sort the right subarray is $$$\lceil \frac{a_i}{a_{i+1}} \rceil-1$$$. Do not forget to update the value of $$$a_i$$$ after the transition from $$$a_i$$$ to $$$a_{i-1}$$$.

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