For the first subtask, we can find the intimidation value of a subarray by looping through it, and increasing a counter if the current element is larger than all elements before it.

Time Complexity: ~\mathcal O(QN)~

For the second subtask, we can sort the queries by descending ~l~ value. As we loop through ~A~ from right to left, we can keep a monotonically increasing stack to keep track of which values to the right of the current value are greater than it. We can maintain this stack in amortized ~\mathcal O(1)~ time. The intimidation value of an array is thus the number of elements in said stack which are less than the current query's ~r~ value. This can be computed in ~\mathcal O(\log N)~ time with a Binary Indexed Tree.

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