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.Scanner;
import java.io.OutputStream;
import java.io.IOException;
import java.io.PrintWriter;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 * @author Vaibhav Mittal
 */
public class Main {
	public static void main(String[] args) {
		InputStream inputStream = System.in;
		OutputStream outputStream = System.out;
		Scanner in = new Scanner(inputStream);
		PrintWriter out = new PrintWriter(outputStream);
		TaskD solver = new TaskD();
		solver.solve(1, in, out);
		out.close();
	}
}

class TaskD {
	public void solve(int testNumber, Scanner in, PrintWriter out) {
        int n = in.nextInt();
        int m = in.nextInt();
        int[] a = new int[n];
        for (int i = 0; i < n; ++i)
            a[i] = in.nextInt();
        int[] l = new int[m];
        int[] r = new int[m];
        for (int i = 0; i < m; ++i) {
            l[i] = in.nextInt();
            r[i] = in.nextInt();
        }
        int[] count = new int[n + 1];
        for (int i = 0; i < n; ++i)
            if (a[i] <= n)
                ++count[a[i]];
        int[] res = new int[m];
        for (int i = 1; i <= n; ++i) {
            if (count[i] < i) continue;
            int[] dp = new int[n + 1];
            for (int j = 0; j < n; ++j)
                dp[j + 1] = dp[j] + (a[j] == i ? 1 : 0);
            for (int j = 0; j < m; ++j)
                if (dp[r[j]] - dp[l[j] - 1] == i)
                    ++res[j];
        }
        for (int i = 0; i < m; ++i)
            out.println(res[i]);
	}
}

