The Little Elephant loves playing with arrays. He has array a, consisting of n positive integers, indexed from 1 to n. Let's denote the number with index i as ai. 

Additionally the Little Elephant has m queries to the array, each query is characterised by a pair of integers lj and rj (1 ≤ lj ≤ rj ≤ n). For each query lj, rj the Little Elephant has to count, how many numbers x exist, such that number x occurs exactly x times among numbers alj, alj + 1, ..., arj.

Help the Little Elephant to count the answers to all queries.

Input

The first line contains two space-separated integers n and m (1 ≤ n, m ≤ 105) — the size of array a and the number of queries to it. The next line contains n space-separated positive integers a1, a2, ..., an (1 ≤ ai ≤ 109). Next m lines contain descriptions of queries, one per line. The j-th of these lines contains the description of the j-th query as two space-separated integers lj and rj (1 ≤ lj ≤ rj ≤ n).

Output

In m lines print m integers — the answers to the queries. The j-th line should contain the answer to the j-th query.

Examples

Input

7 2
3 1 2 2 3 3 7
1 7
3 4


Output

3
1
import java.io.*;
import java.util.*;
public class Main{
public static void main(String []args){
	Scanner in=new Scanner(System.in);
	final int MAXN=100005;
	int n,m,a[],b[],st[],en[],sum[],ans[];
	n=in.nextInt();
	m=in.nextInt();
	a=new int[MAXN];
	b=new int[MAXN];
	ans=new int[MAXN];
	st=new int[MAXN];
	en=new int[MAXN];
	for(int i=1;i<=n;i++){
		a[i]=in.nextInt();
		if(a[i]<=n)b[a[i]]++;
	}
	for(int i=1;i<=m;i++){
		st[i]=in.nextInt();
		en[i]=in.nextInt();
	}
	for(int i=1;i<=n;i++){
		if(i<=b[i]){
			sum=new int[MAXN];
			for(int j=1;j<=n;j++){
				if(a[j]==i){
					sum[j]=sum[j-1]+1;
				}else{
					sum[j]=sum[j-1];
				}
			}
			for(int j=1;j<=m;j++){
				if(sum[en[j]]-sum[st[j]-1]==i)
					ans[j]++;
			}
		}
	}
	for(int i=1;i<=m;i++){
		System.out.println(ans[i]);
	}
}}