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.ArrayList;
import java.util.Arrays;
import java.util.Scanner;

/**
 * User: serparamon
 */
public class Codeforces_2012_08_31_B {

    public static void main(String[] args) throws IOException {
        StreamTokenizer in = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
        PrintWriter out = new PrintWriter(System.out);
        in.nextToken();
        int n = (int) in.nval;
        in.nextToken();
        int m = (int) in.nval;
        int[] k = new int[n+1];
        int[] a = new int[n+1];

        for (int i=1; i<=n; i++) {
            in.nextToken();
            a[i] = (int) in.nval;
            if (a[i] <= n)
                k[a[i]]++;
        }

        int[] l = new int[m];
        int[] r = new int[m];
        int[] ans = new int[m];
        for (int i=0; i<m; i++) {
            in.nextToken();
            l[i] = (int) in.nval;
            in.nextToken();
            r[i] = (int) in.nval;
        }
        
        for (int z=1; z<=n; z++)
            if (k[z] >= z) {
                int[] s = new int[n+1];
                for (int i=0; i<n; i++) 
                    s[i+1] = s[i] + (a[i+1]==z ? 1 : 0);
               
                for (int i=0; i<m; i++)
                    if (s[r[i]] - s[l[i]-1] == z)
                        ans[i]++;
            }
        
        for (int i=0; i<m; i++)
            out.println(ans[i]);

        out.flush();
        out.close();
    }

}
