Problem: Touko's favorite sequence of numbers is a permutation a_1, a_2, ..., a_n of 1, 2, ..., n, and she wants some collection of permutations that are similar to her favorite permutation.

She has a collection of q intervals of the form [l_i, r_i] with 1 ≤ l_i ≤ r_i ≤ n. To create permutations that are similar to her favorite permutation, she coined the following definition:

  * A permutation b_1, b_2, ..., b_n allows an interval [l', r'] to holds its shape if for any pair of integers (x, y) such that l' ≤ x < y ≤ r', we have b_x < b_y if and only if a_x < a_y. 
  * A permutation b_1, b_2, ..., b_n is k-similar if b allows all intervals [l_i, r_i] for all 1 ≤ i ≤ k to hold their shapes. 



Yuu wants to figure out all k-similar permutations for Touko, but it turns out this is a very hard task; instead, Yuu will encode the set of all k-similar permutations with directed acylic graphs (DAG). Yuu also coined the following definitions for herself:

  * A permutation b_1, b_2, ..., b_n satisfies a DAG G' if for all edge u → v in G', we must have b_u < b_v. 
  * A k-encoding is a DAG G_k on the set of vertices 1, 2, ..., n such that a permutation b_1, b_2, ..., b_n satisfies G_k if and only if b is k-similar. 



Since Yuu is free today, she wants to figure out the minimum number of edges among all k-encodings for each k from 1 to q.

Input

The first line contains two integers n and q (1 ≤ n ≤ 25 000, 1 ≤ q ≤ 100 000).

The second line contains n integers a_1, a_2, ..., a_n which form a permutation of 1, 2, ..., n.

The i-th of the following q lines contains two integers l_i and r_i. (1 ≤ l_i ≤ r_i ≤ n).

Output

Print q lines. The k-th of them should contain a single integer — The minimum number of edges among all k-encodings.

Examples

Input


4 3
2 4 1 3
1 3
2 4
1 4


Output


2
4
3


Input


8 4
3 7 4 8 1 5 2 6
3 6
1 6
3 8
1 8


Output


3
5
9
7


Input


10 10
10 5 1 2 7 3 9 4 6 8
2 2
4 5
6 8
4 10
4 4
2 7
2 2
7 8
3 7
2 10


Output


0
1
3
6
6
9
9
9
9
8

Note

For the first test case:

  * All 1-similar permutations must allow the interval [1, 3] to hold its shape. Therefore, the set of all 1-similar permutations is \{[3, 4, 2, 1], [3, 4, 1, 2], [2, 4, 1, 3], [2, 3, 1, 4]\}. The optimal encoding of these permutations is <image>
  * All 2-similar permutations must allow the intervals [1, 3] and [2, 4] to hold their shapes. Therefore, the set of all 2-similar permutations is \{[3, 4, 1, 2], [2, 4, 1, 3]\}. The optimal encoding of these permutations is <image>
  * All 3-similar permutations must allow the intervals [1, 3], [2, 4], and [1, 4] to hold their shapes. Therefore, the set of all 3-similar permutations only includes [2, 4, 1, 3]. The optimal encoding of this permutation is <image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>

int main() {
	using namespace std;
	ios_base::sync_with_stdio(false), cin.tie(nullptr);

	int N, Q; cin >> N >> Q;
	vector<int> A(N); for (auto& a : A) { cin >> a; a--; }
	vector<int> leftmost(N); iota(leftmost.begin(), leftmost.end(), 0);
	struct vert_dat {
		pair<int, int> prv_pred;
		pair<int, int> prv_succ;
		pair<int, int> nxt_pred;
		pair<int, int> nxt_succ;
	};
	vector<vert_dat> verts(N);
	for (int i = 0; i < N; i++) {
		verts[i].prv_pred = {-1, -1};
		verts[i].prv_succ = {N, -1};
		verts[i].nxt_pred = {-1, -1};
		verts[i].nxt_succ = {N, -1};
	}

	int ans = 0;
	for (int q = 0; q < Q; q++) {
		int L, R; cin >> L >> R; L--;
		int b = R;
		while (leftmost[b-1] > L) b--;
		for (; b < R; b++) {
			assert(b > L && leftmost[b-1] <= L);
			while (leftmost[b] > L) {
				leftmost[b]--;
				int a = leftmost[b];
				// insert the a-b constraint
				if (A[a] < A[b] && A[b] < verts[a].nxt_succ.first) {
					if (verts[a].nxt_succ.second != -1) {
						int c = verts[a].nxt_succ.second;
						verts[c].prv_pred.second = -1;
						ans--;
					}
					verts[a].nxt_succ = {A[b], b};
					if (verts[b].prv_pred.second != -1) {
						int c = verts[b].prv_pred.second;
						verts[c].nxt_succ.second = -1;
						ans--;
					}
					verts[b].prv_pred = {A[a], a};
					ans++;
				} else if (A[a] > A[b] && A[b] > verts[a].nxt_pred.first) {
					if (verts[a].nxt_pred.second != -1) {
						int c = verts[a].nxt_pred.second;
						verts[c].prv_succ.second = -1;
						ans--;
					}
					if (verts[b].prv_succ.second != -1) {
						int c = verts[b].prv_succ.second;
						verts[c].nxt_pred.second = -1;
						ans--;
					}
					verts[a].nxt_pred = {A[b], b};
					verts[b].prv_succ = {A[a], a};
					ans++;
				} else {
					continue;
				}
			}
		}
		cout << ans << '\n';
	}

	return 0;
}