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 f[100010][500];
int a[100010], t[100010], l[500];
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  memset(t, 0, sizeof(t));
  memset(f, 0, sizeof(f));
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    if (a[i] <= n) t[a[i]]++;
  }
  int cnt = 0;
  for (int i = 1; i <= n; i++) {
    if (t[i] >= i) l[cnt++] = i;
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < cnt; j++) f[i][j] = f[i - 1][j] + (l[j] == a[i]);
  }
  for (int i = 0; i < m; i++) {
    int ans = 0;
    int left, right;
    scanf("%d%d", &left, &right);
    for (int j = 0; j < cnt; j++) {
      if (f[right][j] - f[left - 1][j] == l[j]) ans++;
    }
    cout << ans << endl;
  }
  return 0;
}
