#include<cstdio>
#define N 25005

int n,q,a[N];

int l[N],r[N],pre[N][2],suf[N][2];

int ans;

int main(){
	scanf("%d%d",&n,&q);
	for(int i=1;i<=n;i++)
		scanf("%d",&a[i]);
	a[0]=n+1;
	for(int i=1;i<=n;i++){
		l[i]=i-1,r[i]=i+1;
		pre[i][1]=suf[i][1]=0;
		pre[i][0]=suf[i][0]=n+1;
	}
	while(q--){
		int L,R;
		scanf("%d%d",&L,&R);
		for(int i=L;i<=R;i++){
			while(l[i]>=L){
				int o=a[l[i]]>a[i];
				if((a[l[i]]>a[pre[i][o]])^o){
					ans+=(i==suf[l[i]][o^1])-(i==suf[pre[i][o]][o^1]);
					pre[i][o]=l[i];
				}
				l[i]--;
			}
			while(r[i]<=R){
				int o=a[r[i]]>a[i];
				if((a[r[i]]>a[suf[i][o]])^o){
					ans+=(i==pre[r[i]][o^1])-(i==pre[suf[i][o]][o^1]);
					suf[i][o]=r[i];
				}
				r[i]++;
			}
		}
		printf("%d\n",ans);
	}
}
### Prompt

Your task is to create a CPP solution to 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<cstdio>
#define N 25005

int n,q,a[N];

int l[N],r[N],pre[N][2],suf[N][2];

int ans;

int main(){
	scanf("%d%d",&n,&q);
	for(int i=1;i<=n;i++)
		scanf("%d",&a[i]);
	a[0]=n+1;
	for(int i=1;i<=n;i++){
		l[i]=i-1,r[i]=i+1;
		pre[i][1]=suf[i][1]=0;
		pre[i][0]=suf[i][0]=n+1;
	}
	while(q--){
		int L,R;
		scanf("%d%d",&L,&R);
		for(int i=L;i<=R;i++){
			while(l[i]>=L){
				int o=a[l[i]]>a[i];
				if((a[l[i]]>a[pre[i][o]])^o){
					ans+=(i==suf[l[i]][o^1])-(i==suf[pre[i][o]][o^1]);
					pre[i][o]=l[i];
				}
				l[i]--;
			}
			while(r[i]<=R){
				int o=a[r[i]]>a[i];
				if((a[r[i]]>a[suf[i][o]])^o){
					ans+=(i==pre[r[i]][o^1])-(i==pre[suf[i][o]][o^1]);
					suf[i][o]=r[i];
				}
				r[i]++;
			}
		}
		printf("%d\n",ans);
	}
}
```