Solving this problem efficiently requires precomputing various useful values, and then using them to answer the questions.

Let's consider each of the ~N~ button sequences. Assuming that sequence ~i~ is executed with the turbolift starting at floor ~0~, let ~len[i]~ be the length of the sequence, ~delta[i]~ be the turbolift's final floor, ~maxP[i][j]~ be the highest floor reached at any point during the first ~j~ button presses of the sequence, ~minP[i][j]~ be the lowest floor reached during the first ~j~ button presses, ~maxS[i]~ be the highest floor reached at any point during the whole sequence (equal to ~maxP[i][len[i]]~), and ~minS[i]~ be the lowest floor reached (equal to ~minP[i][len[i]]~). These values can all be easily computed by simulating the sequence's button presses in ~\mathcal O(L[i])~ time, which is sufficient as the sum of the sequences' lengths is at most ~200\,000~.

Next, let's repeat a similar process over the entire sequence of ~M~ button sequences. Let ~len_2[i]~ be the total length of the first ~i~ executed sequences, ~delta_2[i]~ be the turbolift's final floor after the first ~i~ sequences, ~max_2[i]~ be the highest floor reached at any point during the first ~i~ sequences, and ~min_2[i]~ be the lowest floor reached during the first ~i~ sequences. These values can similarly be easily computed in ~\mathcal O(M)~ time, with the help of the precomputed ~len~, ~delta~, ~maxS~, and ~minS~ values. Note that ~len_2[0] = delta_2[0] = max_2[0] = min_2[0] = 0~.

Finally, we're set up to answer the questions efficiently. Let's say we're interested in the first ~b~ button presses. The ~b~-th button press must occur during some sequence ~i~ ~(1 \le i \le M)~ – in particular, during the first sequence ~i~ such that ~b \le len_2[i]~. Since the values ~len_2[0 \dots M]~ are increasing, we can use binary search to determine the value of ~i~ in ~\mathcal O(\log M)~ time. Now, since the first ~i-1~ button sequences have been completed before the ~b~-th button press, the highest floor reached during the first ~b~ button presses is at least ~max_2[i-1]~. However, the first ~b-len_2[i-1]~ button presses of button sequence ~S_i~ were additionally executed after that, which is where more of our precomputed values come into play – the answer must be ~\max(max_2[i-1], delta_2[i-1]+maxP[S_i][b-len_2[i-1]])~. Similarly, the minimum floor reached is ~\min(min_2[i-1], delta_2[i-1]+minP[S_i][b-len_2[i-1]])~.

The time complexity of this algorithm is ~\mathcal O(\sum_{i=1}^N |B_i| + M + Q \log M)~.