It is obvious that to prolong the game as much as possible, alternating players will move the dot in alternating directions. And conveniently speaking, each turn will only increase the absolute value of the dot's coordinate by $$$1$$$. Proof of this is trivial enough to be left as a mathematical task for readers.

Thus, the player doing the last turn can be determined by the parity of $$$n$$$: the one making the dot's absolute coordinate $$$n+1$$$ would be such player, and thus it'd be Kosuke if $$$n$$$ is odd, and Sakurako otherwise.

Ignore the square. We can see that an operation (if done optimally) will affect all cells within a NW-SE diagonal, or in other words, all cells $$$(i, j)$$$ that have the same $$$i - j$$$ value. Thus, the answer will be the sum of the minimum value's negative for each diagonal (capped at $$$0$$$, so that we won't do anything to those diagonals that are already non-negative).

The easiest way to implement it is to loop $$$i - j$$$ in range $$$[0, 2n-2]$$$ first, then start picking $$$i$$$ in range $$$[0, n-1]$$$ and calculate $$$j$$$ accordingly. Remember to skip out-of-bound $$$j$$$.

Since the swapping operation only allowed indices that are symmetric by the central point, we could loop from one side to the middle to avoid redundancy. Without loss of generality, we choose indices $$$i$$$ from the left side of the array.

For each $$$i$$$, we want to see if either $$$i$$$ or $$$n+1-i$$$ is contributing into disturbance with an element further into the center. It's a matter of casework there:

• If $$$i = n+1-i$$$, ignore as there is nothing inside them.
• Similarly if $$$i+1 = n+1-i$$$; i.e., the symmetrical indices are adjacent, but you'd have to check once if they are equal. If so, they will add $$$1$$$ to the disturbance value.
• If $$$a_i = a_{n+1-i}$$$ and $$$a_{i+1} = a_{n-i}$$$, add $$$2$$$, since both indices will contribute into disturbance regardless of how you swap them.
• If $$$a_i \ne a_{n+1-i}$$$ and $$$a_{i+1} \ne a_{n-i}$$$, add $$$0$$$, since you could swap the indices in a way to ensure the inner-adjacent pairs never have the same value.
• Otherwise, add $$$1$$$ if $$$a_i = a_{i+1}$$$ or $$$a_i = a_{n-i}$$$.

The code does a slightly different casework, but the core idea is the same.

We can see that we'll greedily take the segments from left to right for the optimal outcome.

The idea of prefix sum could help here. Let's denote $$$p_r$$$ as the sum of the prefix until index $$$r$$$ of $$$a$$$, for each $$$r$$$ we're iterating, we want to check if there exists an index $$$l \lt r$$$ such as $$$p_l = p_r$$$.

Thus, we can build a BST (as a set in STL C++), initiated with a single value $$$0$$$. We keep adding up $$$p_r$$$ in the process, and if at any $$$p_r$$$ we find that its value is already in the set, add $$$1$$$ to the final answer, and reset the set into initial state (having only a $$$0$$$).

Re-shape the permutation into a directed graph where a directed edge will be formed from node $$$i$$$ to node $$$p_i$$$. We see that any graph formed this way will be a cluster of simple cycles. A simple permutation, as stated, is a permutation that when converted, it only consists of simple cycles with at most $$$2$$$ nodes each.

We can do a pseudo DFS to easily detect all the cycles (they are pairwise-non-intersecting). With each cycle of size $$$k$$$, we'll need $$$\lfloor \frac{k-1}{2} \rfloor$$$ operations. The catch is that each operation will ensure that one cycle of size $$$2$$$ is made (except the final one, which will break into one such cycle and another cycle of size at most $$$2$$$).

I honestly don't know how to prove this one. But at any rate, there are a few observations that you could trust intuitively to proceed:

• The Fibonacci numbers divisible by $$$k$$$ will always come periodically (i.e., the first number is at index $$$m$$$, the second at index $$$2m$$$, and so on).
• The period size is of linear order over $$$k$$$.

Thus, an $$$\mathcal{O}(k)$$$ solution is possible to find the period size $$$m$$$, then the answer is $$$nm \mod (10^9 + 7)$$$.

A few quick observations:

• If moving upwards consume $$$10^{100}$$$ stamina for each step, then answer for starting vertex $$$v_i$$$ is obviously the deepest children of the subtree rooted at $$$v_i$$$.
• If $$$k_i = 1$$$, denote $$$p$$$ as direct parent of $$$v_i$$$ (we assume that $$$v_i \ne 1$$$ i.e., there exists a node $$$p$$$), then the answer is the maximal distance from $$$v_i$$$ between the deepest children of the subtree rooted at $$$v_i$$$ and the deepest children of the subtree rooted at $$$p$$$ minus subtree rooted at $$$v_i$$$.

We can treat the second part as a reroot: if initially root is $$$p$$$ and then it's handed down to $$$v_i$$$, the distance of all nodes in subtree $$$v_i$$$ drop by $$$1$$$, but the distance of all nodes in subtree $$$p$$$ minus subtree $$$v_i$$$ increase by $$$1$$$. This can be rephrase as: increase all distance by $$$1$$$, drop subtree $$$v_i$$$ by $$$2$$$.

Thus, we can solve the queries offline.

For each query $$$(v_i, k_i)$$$, the scanning scope (i.e., reachable nodes) will be the subtree rooted at the ancestor exactly $$$k_i$$$ steps away from $$$v_i$$$ (or $$$1$$$ if such ancestor doesn't exist).

Here, we assume that you have already calculated all the depth of nodes in the case of root being $$$1$$$. Then, perform a DFS. When reaching each node $$$x$$$, add $$$1$$$ to all depths and subtract $$$2$$$ to the depths under subtree $$$x$$$. Then, solve every query $$$(x, k_i)$$$ simply by finding the maximal depth among nodes under subtree $$$p$$$, with $$$p$$$ being the ancestor exactly $$$k_i$$$ steps away from $$$x$$$ (or $$$1$$$).

Maintaining depths like this can be done with a post-Euler-tour segment tree.