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.*;
import java.io.*;
public class b{
	static Scanner in  = new Scanner(System.in);
	static PrintWriter out = new PrintWriter(System.out);
	int n, m, b;
	int[] a;
	int[][] q;
	int sz = 1000010;
	int[] freq = new int[100100];
	int curr = 0;
	void add(int ind){
		if(a[ind] > sz) return;
		freq[a[ind]]++;
		if(freq[a[ind]] == a[ind]) curr ++ ;
		if(freq[a[ind]] == a[ind] + 1) curr --;
	}
	
	void remove(int ind){
		if(a[ind] > sz) return;
		freq[a[ind]] -- ;
		if(freq[a[ind]] == a[ind]) curr ++;
		if(freq[a[ind]] == a[ind] -1 ) curr--;
	}
	void solve(){
		n = in.nextInt();
		m = in.nextInt();
		a = new int[n+1];
		for(int i=1; i<=n; i++)a[i] = in.nextInt();
		q = new int[m][3];
		for(int i=0; i<m; i++){
			q[i][0] = in.nextInt();
			q[i][1] = in.nextInt();
			q[i][2] = i;
		}
		final int bs= (int)Math.sqrt(n);
		Arrays.sort(q, new Comparator<int[]>(){
			public int compare(int[] a, int[] b){
				if(a[0]/bs != b[0]/bs) return a[0] - b[0];
				return a[1] - b[1];
			}
		});
		int[] ans = new int[m];
		int l = 1, r = 0;
		for(int i=0; i<m; i++)
		{
			int L = q[i][0], R = q[i][1];
			while(r < R) add(++r);
			while(r > R) remove(r--);
			while(l > L) add(--l);
			while(l < L) remove(l++);
			ans[q[i][2]] = curr;

		}
		for(int i=0; i<m; i++) out.printf("%d\n", ans[i]);

	}

	public static void main(String[] args){
		b m = new b();
		m.solve();
		out.flush();
		out.close();
	}
}