#include<bits/stdc++.h>
using namespace std;
const int N=25005;
int n,q,s,p[N],a[N],l1[N],l2[N],r1[N],r2[N];
void la(int x,int y){r2[x]=y,l1[y]=x,s++;}
void lb(int x,int y){r1[x]=y,l2[y]=x,s++;}
void ca(int x,int y){s-=(r2[x]==y&&l1[y]==x);}
void cb(int x,int y){s-=(r1[x]==y&&l2[y]==x);}
void sol(int x,int y)
{
	if(p[x]<p[y]){if(p[x]>p[l1[y]])ca(x,r2[x]),ca(l1[y],y),la(x,y);}
	else{if(p[y]>p[r1[x]])cb(x,r1[x]),cb(l2[y],y),lb(x,y);}
}
int main()
{
	scanf("%d%d",&n,&q);
	for(int i=1;i<=n;i++)
		scanf("%d",&p[i]),a[i]=i;
	while(q--)
	{
		int l,r,i;
		scanf("%d%d",&l,&r);
		for(int i=l;i<=r;i++)
		{
			while(a[i]>l)
			{
				a[i]--;
				sol(a[i],i);
			}
		}
		printf("%d\n",s);
	}
	return 0;
}
### Prompt

Please formulate 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<bits/stdc++.h>
using namespace std;
const int N=25005;
int n,q,s,p[N],a[N],l1[N],l2[N],r1[N],r2[N];
void la(int x,int y){r2[x]=y,l1[y]=x,s++;}
void lb(int x,int y){r1[x]=y,l2[y]=x,s++;}
void ca(int x,int y){s-=(r2[x]==y&&l1[y]==x);}
void cb(int x,int y){s-=(r1[x]==y&&l2[y]==x);}
void sol(int x,int y)
{
	if(p[x]<p[y]){if(p[x]>p[l1[y]])ca(x,r2[x]),ca(l1[y],y),la(x,y);}
	else{if(p[y]>p[r1[x]])cb(x,r1[x]),cb(l2[y],y),lb(x,y);}
}
int main()
{
	scanf("%d%d",&n,&q);
	for(int i=1;i<=n;i++)
		scanf("%d",&p[i]),a[i]=i;
	while(q--)
	{
		int l,r,i;
		scanf("%d%d",&l,&r);
		for(int i=l;i<=r;i++)
		{
			while(a[i]>l)
			{
				a[i]--;
				sol(a[i],i);
			}
		}
		printf("%d\n",s);
	}
	return 0;
}
```