For ~30\%~ of the points, brute-forcing over the ~\binom{2N}{N}~ possible card configurations passes.

Time Complexity: ~\mathcal O(NQ\cdot \binom{2N}{N}+N\cdot 2^{2N})~

Consider a card with value ~v~ placed at the ~i^\text{th}~ position of the first row. Then it can be seen that the first ~i-1~ cards of the first row and the first ~v-i~ cards of the second row must contain all the numbers from ~1~ to ~v-1~. The situation is similar if the card were in the second row.

For the next ~30\%~ of the points, we use the above observation. Let ~u~ be the highest value lower than ~v~ which has been revealed and ~w~ be the lowest value higher than ~v~ which has been revealed. (If ~v~ is already revealed, then the answer is clearly ~1~.) Since we know the region where all the cards from ~1~ to ~u~ are, and we know the region where all the cards from ~1~ to ~w-1~ are, we can see that any card with value between ~u~ and ~w~ must be in a certain region, specifically the union of two subarrays. Additionally, the positions of the cards with value less than ~u~ and the cards with value greater than ~w~ do not affect ~v~'s position.

So it is enough to consider these two subarrays containing all cards with value between ~u~ and ~w~. It is not hard to see that the cards which can have value ~v~ form two subarrays. The boundary points of these subarrays can be found by placing the values in sorted orders and can be directly computed from knowing the positions of ~u~ and ~w~. Alternatively, use the fact that the region of cards with value from ~1~ to ~v~ must contain the region of cards from value ~1~ to ~u~ and similarly with ~w~ to find the positions where ~v~ can be.

Time Complexity: ~\mathcal O(NQ)~

For the final subtask, ~u~ and ~w~ can be found in ~\mathcal O(\log N)~ by using a balanced binary search tree.

Time Complexity: ~\mathcal O(N \log N)~