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 main() {
  ios_base::sync_with_stdio(0);
  int n, m;
  scanf("%d %d\n", &n, &m);
  vector<int> v(n);
  vector<pair<int, pair<int, int>>> q(m);
  for (int i = 0; i < n; i++) {
    scanf("%d ", &v[i]);
  }
  for (int i = 0; i < m; i++) {
    int l, r;
    scanf("%d %d", &l, &r);
    q[i] = {i, {l - 1, r - 1}};
  }
  sort(q.begin(), q.end(),
       [](pair<int, pair<int, int>> a, pair<int, pair<int, int>> b) {
         return a.second.second < b.second.second;
       });
  vector<int> res(q.size());
  for (int a = 0, b = 300; a < v.size(); a += 300, b += 300) {
    unordered_map<int, int> map;
    int l = a, r = a - 1;
    int cnt = 0;
    for (auto pp : q) {
      if (pp.second.first < a || pp.second.first >= b) continue;
      while (r < pp.second.second) {
        map[v[++r]]++;
        if (v[r] == map[v[r]]) cnt++;
        if (map[v[r]] - v[r] == 1) cnt--;
      }
      while (l > pp.second.first) {
        map[v[--l]]++;
        if (v[l] == map[v[l]]) cnt++;
        if (map[v[l]] - v[l] == 1) cnt--;
      }
      while (l < pp.second.first) {
        if (v[l] == map[v[l]]) cnt--;
        map[v[l]]--;
        if (v[l] == map[v[l]]) cnt++;
        l++;
      }
      res[pp.first] = cnt;
    }
  }
  for (int e : res) printf("%d\n", e);
}
