I will use 0-based indexing for convenience with mods.

Note that $$$s_{i+1} - s_i = (a_{i+1} + \ldots + a_{i+m}) - (a_i + \ldots + a_{i+m-1}) = a_{i+m} - a_i$$$, so for all $$$i$$$, the value of $$$a_i$$$ is determined by $$$a_{i \bmod m}$$$ and the input array. More formally, $$$a_i = a_{i\bmod m} + c_i$$$ for some constant $$$c_i$$$, and we can pick $$$a_0 \sim a_{m-1}$$$ freely as long as $$$a_0 + a_1 + \ldots + a_{m-1} = s_0$$$.

If $$$r - l + 1 \lt m$$$, then the answer is unbounded. Otherwise, we know that each query is asking for the smallest possible value of $$$\max(a_0 + x_0, a_1 + x_1, \ldots, a_{m-1} + x_{m-1})$$$ where $$$x_i = \underset{l\leq j\leq r \wedge j \equiv i\bmod m}{\max} c_j$$$.

To get the minimum value of the above expression, consider when we can make it $$$\leq t$$$ (think binary searching on answer). We get $$$a_i + x_i \leq t$$$ for $$$0 \leq i \lt m$$$, and summing up all the equations, $$$s_0 + \sum {x_i} \leq mt$$$. Rearranging gives $$$t \geq \left\lceil \frac{s_0 + \sum {x_i}}{m} \right\rceil$$$ which is the answer.

Now, finally, stuff that isn't in the official editorial: we process the queries in order of increasing $$$r$$$. This is nice because if we are currently answering queries with $$$r = j$$$ when $$$c_0$$$ to $$$c_j$$$ have been processed, and we see that $$$c_i \leq c_j$$$ where $$$i \lt j$$$ and $$$i \equiv j \pmod m$$$, then $$$c_i$$$'s contribution to $$$x_{j \bmod m}$$$ will be overshadowed by $$$c_j$$$.

Therefore, we can maintain a monotonic stack for each set of indices that give the same value $$$\bmod m$$$, where the topmost elements have larger index and smaller values. If $$$c_i$$$ and $$$c_j$$$ $$$(i \lt j)$$$ are next to each other in our monotonic stack, then that means if the query left bound $$$l \leq i$$$, it will contribute $$$c_i - c_j$$$ to our answer.

We can use a point-update, range-sum segment tree to maintain this difference array on our monotonic stack. To get $$$\sum {x_i}$$$, we just query the range sum from $$$l$$$ to $$$r$$$ at the moment we have processed $$$c_0$$$ to $$$c_r$$$. It runs in $$$O((n + q) \log n)$$$.