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.*;

public class LittleElphantAndArray {
    public static void main(String args[]) {
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        int m = in.nextInt();
        int[] arr = new int[n];
        int[] freq = new int[n + 1];
        int[] start = new int[m];
        int[] end = new int[m];
        int[] sol = new int[m];
        for (int i = 0; i < n; i++) {
            int tmp = in.nextInt();
            arr[i] = tmp;
            if (tmp <= n) {
                freq[tmp]++;
            }
        }
        for (int i = 0; i < m; i++) {
            int tmp = in.nextInt();
            int tmp2 = in.nextInt();
            start[i] = tmp;
            end[i] = tmp2;
        }
        for (int i = 1; i <= n; i++) {
            if (freq[i] < i) {
                continue;
            }
            int tfreq[] = new int[n + 1];
            for (int j = 1; j <= n; j++) {
                tfreq[j] = tfreq[j - 1];
                if (arr[j - 1] == i)
                    tfreq[j]++;
            }
            for (int j = 0; j < m; j++) {
                if (tfreq[end[j]] - tfreq[start[j] - 1] == i) {
                    sol[j]++;
                }
            }
        }
        for (int i = 0; i < m; i++) {
            System.out.println(sol[i]);
        }
    }
}