#include<bits/stdc++.h>
#define cs const
#define pb push_back
using namespace std;

cs int N = 25005;

int n, m, a[N];
int lu[N], ld[N], ru[N], rd[N];
int lp[N], rp[N], ans;
void upt(int l, int r) {
	for(int i = l; i <= r; i++) {
		for(int &j = lp[i]; j >= l; j--) {
			if(a[j] > a[i]) {
				int &k = lu[i]; 
				if(a[j] < a[k]) {
					ans += rd[j] == i;
					ans -= rd[k] == i;
					k = j;
				}
			}
			else {
				int &k = ld[i];
				if(a[j] > a[k]) {
					ans += ru[j] == i; 
					ans -= ru[k] == i;
					k = j;
				}
			}
		}
		for(int &j = rp[i]; j <= r; j++) {
			if(a[j] > a[i]) {
				int &k = ru[i]; 
				if(a[j] < a[k]) {
					ans += ld[j] == i;
					ans -= ld[k] == i;
					k = j;
				}
			}
			else {
				int &k = rd[i];
				if(a[j] > a[k]) {
					ans += lu[j] == i; 
					ans -= lu[k] == i;
					k = j;
				}
			}
		}
	}
}
int main() {
	#ifdef FSYo
	freopen("1.in", "r", stdin);
	#endif
	cin >> n >> m;
	for(int i = 1; i <= n; i++)
		scanf("%d", &a[i]);
	a[n + 1] = n + 1; 
	for(int i = 1; i <= n; i++) {
		lu[i] = ru[i] = n + 1;
		lp[i] = i - 1, rp[i] = i + 1;
	}
	for(int i = 1, l, r; i <= m; i++) {
		scanf("%d%d", &l, &r);
		upt(l, r), cout << ans << '\n';
	}
	return 0;
}
### Prompt

Create a solution in CPP for the following 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>

### Response

```cpp
#include<bits/stdc++.h>
#define cs const
#define pb push_back
using namespace std;

cs int N = 25005;

int n, m, a[N];
int lu[N], ld[N], ru[N], rd[N];
int lp[N], rp[N], ans;
void upt(int l, int r) {
	for(int i = l; i <= r; i++) {
		for(int &j = lp[i]; j >= l; j--) {
			if(a[j] > a[i]) {
				int &k = lu[i]; 
				if(a[j] < a[k]) {
					ans += rd[j] == i;
					ans -= rd[k] == i;
					k = j;
				}
			}
			else {
				int &k = ld[i];
				if(a[j] > a[k]) {
					ans += ru[j] == i; 
					ans -= ru[k] == i;
					k = j;
				}
			}
		}
		for(int &j = rp[i]; j <= r; j++) {
			if(a[j] > a[i]) {
				int &k = ru[i]; 
				if(a[j] < a[k]) {
					ans += ld[j] == i;
					ans -= ld[k] == i;
					k = j;
				}
			}
			else {
				int &k = rd[i];
				if(a[j] > a[k]) {
					ans += lu[j] == i; 
					ans -= lu[k] == i;
					k = j;
				}
			}
		}
	}
}
int main() {
	#ifdef FSYo
	freopen("1.in", "r", stdin);
	#endif
	cin >> n >> m;
	for(int i = 1; i <= n; i++)
		scanf("%d", &a[i]);
	a[n + 1] = n + 1; 
	for(int i = 1; i <= n; i++) {
		lu[i] = ru[i] = n + 1;
		lp[i] = i - 1, rp[i] = i + 1;
	}
	for(int i = 1, l, r; i <= m; i++) {
		scanf("%d%d", &l, &r);
		upt(l, r), cout << ans << '\n';
	}
	return 0;
}
```