It is not necessary to use a segment tree or a sparse table to solve this problem. These approaches are overkill and run quite slowly. Binary search is also not necessary, and it increases execution time.

The first observation is that ~\text{anything} \operatorname{NOR} 1 = 0~, where ~\text{anything}~ can be any expression. In particular, if ~A_y = 1~, then the answer to the query ~x\ y~ is always ~0~.

To solve a query, it is only necessary to get the largest ~k~ that satisfies both ~k \le y~ and ~A_k=1~. In case ~A_1 \dots A_y~ are all 0's, let ~k=0~. This information can be preprocessed and stored in an array in ~\mathcal{O}(N)~ time (linear time). Now there are ~3~ cases to consider:

• ~k<x~: In this case, all integers are equal to ~0~.
• ~k=x~: In this case, only the first integer is equal to ~1~. The rest are equal to ~0~.
• ~k>x~: In this case, ~(A_x \operatorname{NOR} A_{x+1} \operatorname{NOR} \dots \operatorname{NOR} A_k) = 0~, because ~A_k=1~ and ~\text{anything} \operatorname{NOR} 1 = 0~.

A query can be solved in ~\mathcal{O}(1)~ (constant time). Make sure to print the correct answer for each case.

Time Complexity: ~\mathcal{O}(N+Q)~