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 fn[100010] = {0};
inline void upd(int i, int val) {
  for (int idx = i; idx <= 100010; idx += (idx & -idx)) fn[idx] += val;
}
inline int gg(int e) {
  int res = 0;
  for (int idx = e; idx; idx -= (idx & -idx)) res += fn[idx];
  return res;
}
int in[100010];
vector<pair<int, int> > qr[100010];
vector<int> ans;
vector<int> aqq[100010];
int main() {
  int n, m;
  cin >> n >> m;
  ans.resize(m);
  for (int i = 1; i < n + 1; i += 1) {
    cin >> in[i];
    if (in[i] > n) in[i] = -1;
  }
  for (int i = 0; i < m; i += 1) {
    int s, e;
    cin >> s >> e;
    qr[e].push_back(make_pair(i, s));
  }
  long long res = 0;
  for (int i = 1; i < n + 1; i += 1) {
    int tt = in[i];
    if (in[i] != -1) {
      aqq[tt].push_back(i);
      if (aqq[tt].size() >= tt) upd(aqq[tt][aqq[tt].size() - tt], 1);
      if (aqq[tt].size() > tt) upd(aqq[tt][aqq[tt].size() - tt - 1], -2);
      if (aqq[tt].size() > tt + 1) upd(aqq[tt][aqq[tt].size() - tt - 2], 1);
    }
    for (int j = 0; j < qr[i].size(); j += 1) {
      int s = qr[i][j].second;
      int ind = qr[i][j].first;
      ans[ind] = gg(i) - gg(s - 1);
    }
  }
  for (int i = 0; i < m; i += 1) cout << ans[i] << endl;
  return 0;
}
