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.util.*;

public class Main
{
	public static void main(String[] args)
	{
		int n, q;
		int[] a, ans, cnt, l, r;
		Scanner cin = new Scanner(System.in);
		
		n = cin.nextInt();  q = cin.nextInt();
		a = new int[n+10];  cnt = new int[n+10];
		ans = new int[q+10];  l = new int[q];  r = new int[q];
		for(int i=1; i<=n; i++)
		{
			a[i] = cin.nextInt();
			if( a[i] <= n )
				cnt[a[i]]++;
		}
		for(int i=0; i<q; i++)
		{
			l[i] = cin.nextInt();
			r[i] = cin.nextInt();
		}
		
		for(int i=1; i<=n; i++)
			if( cnt[i] >= i )
			{
				int[] tmp = new int[n+10];
				for(int j=1; j<=n; j++)
				{
					tmp[j] = tmp[j-1];
					if( a[j] == i )
						tmp[j]++;
				}
				for(int j=0; j<q; j++)
					if( tmp[r[j]]-tmp[l[j]-1] == i)
						ans[j]++;
			}
		
		for(int i=0; i<q; i++)
			System.out.println(ans[i]);
	}

}
