Consider s as combination of bits. Next calculate contribution of each (s[i]*2^i)^2 and (2*SI*sj*2^(i+j)) separately. And some binomial formula to compute these probabilities

Use dp[i][j] — minimal length of a palindrome according to positions in regex.

C: one can see that a sufficiently large power of two satisfies the matrices equality condition. One can also see that every such number is divisible by $$$k$$$ hence $$$k$$$ is a power of two as well. Now we try each power of two until we find a good one.

The easiest way to check the board state after $$$k$$$ operations is to consider inverse operations: before matrix $$$A$$$ the was matrix $$$B$$$ such that $$$B_{ij} = A_{ij} + A_{(i-1)j} + A_{i(j-1)} + A_{(i-1)(j-1)}$$$, and for any $$$k$$$ which is a power of two a board after $$$k$$$ inverse operations looks the same but every $$$1$$$ becomes $$$k$$$.

E: for each b up to 512 create vector x_i -- number of times i appears in the array a. Now multiply all pairs b_i*b_j as polynomials and write to b_(i xor j). Now we have some representation of all pairs (a_i + a_j, b_i xor b_j) Now do it one more time and for each pair add to the answer

E: we'll find cnt[x][y] being equal to the number of sets of 4 indices where sum of $$$a$$$'s is $$$x$$$ and xor of $$$b$$$-s is $$$y$$$. It can be done by a Walsh-Hadamard transform with NTT inside.

C: There is another solution $$$DP[i][j]$$$ — answer for submatrix $$$((0,0), (i, j))$$$
transitions: let $$$x = lcm(DP[i - 1][j], DP[i][j - 1])$$$
clearly to make submatrix(i, j) equal to initial we must do $$$x \cdot k$$$ steps, where $$$k$$$ is integer and $$$k > 0$$$
But if have done $$$x$$$ steps and $$$val[i][j] != initial\ val[i][j]$$$ than next $$$x$$$ steps $$$val[i][j]$$$ will be simply xored, than
$$$DP[i][j] = x$$$ or $$$DP[i][j] = 2 \cdot x $$$
here we can notice that DP is always power of two, than we will store only powers of two
$$$x = max(DP[i - 1][j], DP[i][j - 1])$$$
Thus transition is $$$DP[i][j] = x$$$ or $$$DP[i][j] = x + 1 $$$
but how to determine what transition to make $$$x$$$ or $$$(x + 1)$$$. To make choice we need to calculate $$$f(i, j, 2^x)$$$ — value of $$$(i, j)$$$ element after $$$2^x$$$ steps.
I think it's hardest part of my solution. $$$f(i, j, k) = \sum\limits^{i}_{x = 0} \sum\limits ^{j}_{y=0} f(x, y, k - 1)$$$

So if we recursively open such sums we will reduce to $$$f(i, j, k) = \sum\limits^{i}_{x = 0} \sum\limits ^{j}_{y=0} f(x, y, 0) \cdot ways(x, y, i, j, k)$$$
where $$$ways(x, y, i, j, k)$$$ is how many times element $$$(x, y)$$$ will occurs in sum for $$$f(i, j, k)$$$
It looks like number of ways from $$$(x, y, 0)$$$ to $$$(i, j, k)$$$ in 3D grid but with some limitations of making way.
let $$$dx = i - x$$$ and $$$dy = j - y$$$
Actually $$$ways(x, y, i, j, k)$$$ equals to numbef of ways to make two sequences of size k
$$$a_i > 0$$$ and $$$b_i > 0$$$ such as $$$\sum a_i = dx$$$ and $$$\sum b_i = dy$$$
So it is well known combinatorics problem $$$ways(x, y, i, j, k) = C_{dx + k - 1}^{k - 1} \cdot C_{dy + k - 1}^{k - 1}$$$
Of course every calculation must be computed by (mod 2), and I just noticed that
$$$C_{n + 2^x - 1}^{2^x - 1} (mod\ 2) = 1\ \ if\ \ n\ |\ 2^x\ \ else\ \ 0$$$

This immediately gives you solution $$$O((n \cdot m) ^ 2)$$$
Optimization to $$$O(n \cdot m \cdot \log(n \cdot m))$$$ is easier task.

Implementation

Well, we precalculated answers for all numbers below $$$10^7$$$ and this is all memoization we used, now the number of operations is about $$$10^7 + t\cdot11\cdot10\cdot9\cdot8\cdot7$$$.

DFS the tree to get the DFS order. A vertex's subtree is an interval $$$[l_i,r_i]$$$ of this sequence. When two intervals $$$[l_1,r_1],[l_2,r_2]$$$ have intersection, this pair is counted into the answer. This means $$$l_1\leq r_2$$$ and $$$r_1\geq l_2$$$.

Consider a huge binary matrix $$$g_{i,j}$$$, where $$$g_{i,j}$$$ denotes whether interval $$$[l_i,r_i]$$$ intersects with interval $$$[l_j,r_j]$$$. The answer of query $$$[l,r]$$$ is the sum of a $$$g$$$'s submatrix.

Brute force solution runs in $$$O(n^2)$$$, but we can easily speed it up using bitset. My bitset solution got OK in 1.8 seconds.

I believe $$$O((n + q) \sqrt n \log n)$$$ passes but my coding abilities are too low to squeeze it.

Let's do sqrt decomposition on positions, we will learn to ask $$$get\_subtree(v, l, r)$$$ and have $$$a[i][j]$$$ precalced — the number of vertices from the $$$i$$$-th block that have $$$j$$$ in the subtree.

The first part can be computed with persistent segtree which stores 1 at position $$$i$$$ if vertex $$$i$$$ is in the subtree. Let's build it from the bottom using small-to-large in $$$O(n \log^2 n)$$$.

The second part can be computed with another persistent segtree which stores 1 at position $$$i$$$ if vertex $$$i$$$ is the ancestor. This is built from the top in $$$O(n \log n)$$$. After that we'll iterate through blocks, through all vertices and set $$$a[i][j] = get\_ancestors(j, i * BL, (i + 1) * BL)$$$. This is $$$O(n \sqrt n \log n)$$$.

Finally query works in $$$O(\sqrt n \log n)$$$ with this.

Regard all $$$a_i$$$ as strings and append character 'z' to each string. Then add enough zeroes to the set $$$S$$$ to ensure $$$|S|=\sum |a_i|$$$.

Assume the final answer is $$$x_1,x_2,\dots,x_n$$$. Let's assign digits in the set from $$$0$$$ to $$$9$$$ one by one. Always assign the current digit to the current highest bit of $$$x_k$$$, where $$$a_k$$$ is the string with the smallest lexicographical order. Assume the highest bit of $$$a_k$$$ is $$$A$$$, and now we are assigning digit $$$B$$$. If $$$B>A$$$ then there is no solution. If $$$B=A$$$ then we just erase the highest bit of $$$a_k$$$. If $$$B<A$$$ then we can assign everything to the left part of $$$x_k$$$, so we can just remove this string.

When we removed all the strings, there are no limits at all, then we can assign the left digits easily.