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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;

/**
 * 220B
 * O(n + (n + m)*√n) time
 * θ(n*√n + m) space
 *
 * @author artyom
 */
public class _220B_Faster implements Runnable {
    private BufferedReader in;
    private StringTokenizer tok;

    private Object solve() throws IOException {
        int n = nextInt(), m = nextInt();
        int[] a = new int[n];
        int[] cnt = new int[n + 1];
        List<Integer> ls = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            a[i] = nextInt();
            if (a[i] <= n) {
                if (++cnt[a[i]] == a[i]) {
                    ls.add(a[i]);
                }
            }
        }
        int z = ls.size();
        int[] vs = new int[z]; // Key observation: z <= √2n
        for (int i = 0; i < z; i++) {
            vs[i] = ls.get(i);
        }
        int[][] ps = new int[z][n + 1];
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < z; j++) {
                if (vs[j] == a[i]) {
                    ps[j][i + 1] = ps[j][i] + 1;
                } else {
                    ps[j][i + 1] = ps[j][i];
                }
            }
        }
        StringBuilder sb = new StringBuilder();
        for (int i = 1; i <= m; i++) {
            int l = nextInt(), r = nextInt(), c = 0;
            for (int j = 0; j < z; j++) {
                if (ps[j][r] - ps[j][l - 1] == vs[j]) {
                    c++;
                }
            }
            sb.append(c);
            if (i < m) {
                sb.append('\n');
            }
        }
        return sb;
    }

    //--------------------------------------------------------------
    public static void main(String[] args) {
        new _220B_Faster().run();
    }

    @Override
    public void run() {
        try {
            in = new BufferedReader(new InputStreamReader(System.in));
            tok = null;
            System.out.print(solve());
            in.close();
        } catch (IOException e) {
            System.exit(0);
        }
    }

    private String nextToken() throws IOException {
        while (tok == null || !tok.hasMoreTokens()) {
            tok = new StringTokenizer(in.readLine());
        }
        return tok.nextToken();
    }

    private int nextInt() throws IOException {
        return Integer.parseInt(nextToken());
    }
}