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 r221d {
	public static void main(String args[]) {
		Scanner in = new Scanner(System.in);
		int N = in.nextInt();
		int M = in.nextInt();
		int[] arr = new int[N];
		int[] freq = new int[N+1];
		int[] start = new int[M];
		int[] end = new int[M];
		int[] ans = new int[M];
		for(int i = 0; i < N; i++) {
			int tmp = in.nextInt();
			arr[i] = tmp;
			if(tmp <= N)
				freq[tmp]++;
		}
		
		//System.out.println(Arrays.toString(freq));
		
		for(int i = 0; i < M; i++) {
			int tmp = in.nextInt();
			int tmp2 = in.nextInt();
			start[i] = tmp;
			end[i] = tmp2;
		}
		
		for(int i = 1; i <= N; i++) {
			if(freq[i] < i)
				continue;
			int tfreq[] = new int[N+1];
			for(int j = 1; j <= N; j++) {
				tfreq[j] = tfreq[j-1];
				if(arr[j-1] == i)
					tfreq[j]++;
			}
			
			for(int j = 0; j < M; j++) {
				if(tfreq[end[j]] - tfreq[start[j]-1] == i)
					ans[j]++;
			}
			
			//System.out.printf("%d : %s\n", i, Arrays.toString(tfreq));
		}
		
		for(int i = 0; i < M; i++)
			System.out.println(ans[i]);
		
		
	}
}
