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
#include <bits/stdc++.h>
using namespace std;
int n, a[100500], i, b[100500], m, j, ans, l, r, k, len, kol[500][100500],
    c[1000];
int main() {
  scanf("%d%d\n", &n, &m);
  for (i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    b[i] = a[i];
  }
  scanf("\n");
  sort(b + 1, b + n + 1);
  for (i = 1; i <= n; i++) {
    k++;
    if (b[i] != b[i + 1]) {
      if (k >= b[i]) {
        len++;
        c[len] = b[i];
      }
      k = 0;
    }
  }
  for (i = 1; i <= len; i++) {
    for (j = 1; j <= n; j++) {
      kol[i][j] = kol[i][j - 1];
      if (a[j] == c[i]) kol[i][j]++;
    }
  }
  for (i = 1; i <= m; i++) {
    scanf("%d%d\n", &l, &r);
    ans = 0;
    for (j = 1; j <= len; j++) {
      if (kol[j][r] - kol[j][l - 1] == c[j]) ans++;
    }
    printf("%d\n", ans);
  }
  return 0;
}
