// do a test
using namespace std;
#include <bits/stdc++.h>
#define N 25005
#define M 100005
int n,m;
int a[N];
int lmin[N],rmax[N];
int suc[N][2],pre[N][2];
int ans;
void work(int l,int r){
	for (int i=l;i<=r;++i){
		for (int &j=rmax[i];j<=r;++j)
			if (a[j]>a[i]){
				int &t=suc[i][0];
				if (a[j]<a[t])
					ans+=(pre[j][1]==i)-(pre[t][1]==i),t=j;
			}
			else{
				int &t=pre[i][0];
				if (a[j]>a[t])
					ans+=(suc[j][1]==i)-(suc[t][1]==i),t=j;
			}
		for (int &j=lmin[i];j>=l;--j)
			if (a[j]>a[i]){
				int &t=suc[i][1];
				if (a[j]<a[t])
					ans+=(pre[j][0]==i)-(pre[t][0]==i),t=j;
			}
			else{
				int &t=pre[i][1];
				if (a[j]>a[t])
					ans+=(suc[j][0]==i)-(suc[t][0]==i),t=j;
			}
	}
}
int main(){
	scanf("%d%d",&n,&m);
	for (int i=1;i<=n;++i)
		scanf("%d",&a[i]);
	a[0]=n+1,a[n+1]=0;
	for (int i=1;i<=n;++i){
		lmin[i]=i-1,rmax[i]=i+1;
		suc[i][0]=suc[i][0]=0;
		pre[i][0]=pre[i][1]=n+1;
	}
	for (int i=1;i<=m;++i){
		int l,r;
		scanf("%d%d",&l,&r);
		work(l,r);
		printf("%d\n",ans);
	}
	return 0;
}
### Prompt

Your challenge is to write 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
// do a test
using namespace std;
#include <bits/stdc++.h>
#define N 25005
#define M 100005
int n,m;
int a[N];
int lmin[N],rmax[N];
int suc[N][2],pre[N][2];
int ans;
void work(int l,int r){
	for (int i=l;i<=r;++i){
		for (int &j=rmax[i];j<=r;++j)
			if (a[j]>a[i]){
				int &t=suc[i][0];
				if (a[j]<a[t])
					ans+=(pre[j][1]==i)-(pre[t][1]==i),t=j;
			}
			else{
				int &t=pre[i][0];
				if (a[j]>a[t])
					ans+=(suc[j][1]==i)-(suc[t][1]==i),t=j;
			}
		for (int &j=lmin[i];j>=l;--j)
			if (a[j]>a[i]){
				int &t=suc[i][1];
				if (a[j]<a[t])
					ans+=(pre[j][0]==i)-(pre[t][0]==i),t=j;
			}
			else{
				int &t=pre[i][1];
				if (a[j]>a[t])
					ans+=(suc[j][0]==i)-(suc[t][0]==i),t=j;
			}
	}
}
int main(){
	scanf("%d%d",&n,&m);
	for (int i=1;i<=n;++i)
		scanf("%d",&a[i]);
	a[0]=n+1,a[n+1]=0;
	for (int i=1;i<=n;++i){
		lmin[i]=i-1,rmax[i]=i+1;
		suc[i][0]=suc[i][0]=0;
		pre[i][0]=pre[i][1]=n+1;
	}
	for (int i=1;i<=m;++i){
		int l,r;
		scanf("%d%d",&l,&r);
		work(l,r);
		printf("%d\n",ans);
	}
	return 0;
}
```