check trick number 3 probably something to do with this

Notice that the number of subsets containing elements $$$a_i$$$ and $$$a_j$$$ with $$$\mathrm{sum} \le s$$$ is the same as the number of subsets of the set $$$\{a_1, a_2,\dots, a_n\}\setminus\{a_i, a_j\}$$$ with $$$\mathrm{sum} \le s - a_i - a_j$$$.

First, build an array such that $$$dp[i]$$$ is the number of subsets with $$$\mathrm{sum} = i$$$ for $$$i \in [1, k]$$$. This is a standard dp problem that can be solved in $$$O(nk)$$$.

Now for each query, we would like to remove those two elements from the dp array somehow efficiently. Turns out this is possible: you can remove a single element in $$$O(k)$$$ time. Check out trick 5 from this blog. After the elements have been removed, we have to calculate the sum of some prefix of the dp array. This can be naively done in $$$O(k)$$$. You can then add the two elemwnts back to the knapsack in $$$O(k)$$$ time.

This means that the total time complexity is $$$O(nk + qk) = O((n + q)k)$$$.