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>
int x[100010], c[100010];
bool pss[100010];
int d[600][100010], inv[600];
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &x[i]);
    if (x[i] <= n && ++c[x[i]] == x[i]) pss[x[i]] = 1;
  }
  int cnt = 0;
  for (int i = 1; i <= n; i++) {
    if (!pss[i]) continue;
    for (int j = 1; j <= n; j++) d[cnt][j] = d[cnt][j - 1] + (x[j] == i);
    inv[cnt++] = i;
  }
  while (m--) {
    int l, r, ans = 0;
    scanf("%d%d", &l, &r);
    for (int i = 0; i < cnt; i++)
      if (d[i][r] - d[i][l - 1] == inv[i]) ans++;
    printf("%d\n", ans);
  }
}
