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;
vector<int> can;
int mp[100011];
struct node {
  int l, r;
  int res;
} q[100011];
int num[100011];
int pre[100011];
int main() {
  int n, m;
  while (scanf("%d%d", &n, &m) != EOF) {
    memset(mp, 0, sizeof(mp));
    ;
    for (int i = 1; i <= n; i++) {
      int tt;
      scanf("%d", &tt);
      if (tt > n) continue;
      mp[tt]++;
      if (mp[tt] == tt) can.push_back(tt);
      num[i] = tt;
    }
    for (int i = 1; i <= m; i++) {
      scanf("%d%d", &q[i].l, &q[i].r);
      q[i].res = 0;
    }
    for (int i = 0; i < can.size(); i++) {
      pre[0] = 0;
      for (int j = 1; j <= n; j++) pre[j] = pre[j - 1] + (num[j] == can[i]);
      for (int j = 1; j <= m; j++) {
        if (pre[q[j].r] - pre[q[j].l - 1] == can[i]) q[j].res++;
      }
    }
    for (int i = 1; i <= m; i++) printf("%d\n", q[i].res);
  }
  return 0;
}
