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:
/* https://codeforces.com/blog/entry/89644?#comment-781096 (ecnerwala) */
#include <stdio.h>

#define N	25000

int main() {
	static int aa[N], ll[N], rr[N], lu[N], ld[N], ru[N], rd[N];
	int n, q, ans, i;

	scanf("%d%d", &n, &q);
	for (i = 0; i < n; i++)
		scanf("%d", &aa[i]), aa[i]--;
	for (i = 0; i < n; i++)
		ll[i] = rr[i] = i, lu[i] = ld[i] = ru[i] = rd[i] = -1;
	ans = 0;
	while (q--) {
		int l, r;

		scanf("%d%d", &l, &r), l--, r--;
		for (i = r; ll[i] > l; i--)
			while (ll[i] > l) {
				ll[i]--;
				if (aa[ll[i]] > aa[i]) {
					if (lu[i] == -1 || aa[lu[i]] > aa[ll[i]]) {
						if (lu[i] != -1 && rd[lu[i]] == i)
							ans--;
						lu[i] = ll[i];
						if (rd[lu[i]] == i)
							ans++;
					}
				} else {
					if (ld[i] == -1 || aa[ld[i]] < aa[ll[i]]) {
						if (ld[i] != -1 && ru[ld[i]] == i)
							ans--;
						ld[i] = ll[i];
						if (ru[ld[i]] == i)
							ans++;
					}
				}
			}
		for (i = l; rr[i] < r; i++)
			while (rr[i] < r) {
				rr[i]++;
				if (aa[rr[i]] > aa[i]) {
					if (ru[i] == -1 || aa[ru[i]] > aa[rr[i]]) {
						if (ru[i] != -1 && ld[ru[i]] == i)
							ans--;
						ru[i] = rr[i];
						if (ld[ru[i]] == i)
							ans++;
					}
				} else {
					if (rd[i] == -1 || aa[rd[i]] < aa[rr[i]]) {
						if (rd[i] != -1 && lu[rd[i]] == i)
							ans--;
						rd[i] = rr[i];
						if (lu[rd[i]] == i)
							ans++;
					}
				}
			}
		printf("%d\n", ans);
	}
	return 0;
}