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.*;
import java.math.*;

public class Main {
    static StringTokenizer st;
    static PrintWriter out = new PrintWriter(System.out,false);
    static BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
    public static void main(String[] args) throws Exception {
        int n = nextInt();
        int m = nextInt();
        int[] mas = new int[n+1];
        int[] tmp = new int[n+1];
        int[] rep = new int[n+1];
        int[] l = new int[m];
        int[] r = new int[m];
        int[] ans = new int[m];
        for(int i = 1;i <= n;i++) {
            mas[i] = nextInt();
            if(mas[i] <= n)rep[mas[i]]++;
        }
        for(int i = 0;i < m;i++) {
            l[i] = nextInt();
            r[i] = nextInt();
        }
        for(int i = 1;i <= n;i++) {
            if(rep[i] < i) continue;
            for(int j = 1;j <= n;j++) tmp[j] = tmp[j-1]+(mas[j]==i?1:0);
            for(int j = 0;j < m;j++) if(tmp[r[j]] - tmp[l[j]-1] == i) ans[j]++;
        }
        for(int j = 0;j < m;j++) out.println(ans[j]);
        out.flush();
    }

    public static int nextInt() throws Exception {
        if(st == null || !st.hasMoreTokens()) st = new StringTokenizer(in.readLine());
        return Integer.parseInt(st.nextToken());
    }
}