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 George Marcus
 */
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[] v = new int[N + 1];
        for(int i = 1; i <= N; i++)
            v[i] = in.nextInt();

        int[] left = new int[M];
        int[] right = new int[M];
        int[] ans = new int[M];

        for(int i = 0; i < M; i++) {
            left[i] = in.nextInt();
            right[i] = in.nextInt();
        }

        int[] cnt = new int[N + 1];
        for(int i = 1; i <= N; i++)
            if(v[i] <= N)
                cnt[v[i]]++;

        for(int i = 1; i <= N; i++)
            if(cnt[i] >= i) {
                int[] dp = new int[N + 1];

                dp[0] = 0;
                for(int j = 1; j <= N; j++) {
                    dp[j] = dp[j - 1];
                    if(v[j] == i)
                        dp[j]++;
                }

                for(int j = 0; j < M; j++)
                    if(dp[right[j]] - dp[left[j] - 1] == i)
                        ans[j]++;
            }

        for(int i = 0; i < M; i++)
            out.println(ans[i]);
	}
}

