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;
struct Q {
  int l;
  int r;
  int num;
  Q(int _l, int _r, int _num) : l(_l), r(_r), num(_num) {}
};
int cnt[100100];
int res[100100];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n, m;
  cin >> n >> m;
  vector<int> a(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  vector<int> comp = a;
  sort(comp.begin(), comp.end());
  comp.resize(unique(comp.begin(), comp.end()) - comp.begin());
  for (int i = 0; i < n; i++) {
    a[i] = lower_bound(comp.begin(), comp.end(), a[i]) - comp.begin();
  }
  int K = sqrt(n);
  int B = n / K + 1;
  vector<vector<Q>> blocks(B);
  for (int i = 0; i < m; i++) {
    int l, r;
    cin >> l >> r;
    l--;
    r--;
    blocks[l / K].emplace_back(l, r, i);
  }
  for (int b = 0; b < B; b++) {
    sort(blocks[b].begin(), blocks[b].end(),
         [&](Q x, Q y) { return x.r < y.r; });
  }
  for (int b = 0; b < B; b++) {
    if (blocks[b].empty()) continue;
    memset(cnt, 0, sizeof(cnt));
    int ans = 0;
    for (int i = blocks[b][0].l; i <= blocks[b][0].r; i++) {
      if (cnt[a[i]] == comp[a[i]]) {
        ans--;
      }
      cnt[a[i]]++;
      if (cnt[a[i]] == comp[a[i]]) {
        ans++;
      }
    }
    int pl = blocks[b][0].l;
    int pr = blocks[b][0].r;
    res[blocks[b][0].num] = ans;
    for (int q = 1; q < (int)blocks[b].size(); q++) {
      for (int i = pr + 1; i <= blocks[b][q].r; i++) {
        if (cnt[a[i]] == comp[a[i]]) {
          ans--;
        }
        cnt[a[i]]++;
        if (cnt[a[i]] == comp[a[i]]) {
          ans++;
        }
      }
      pr = blocks[b][q].r;
      for (int i = blocks[b][q].l; i < pl; i++) {
        if (cnt[a[i]] == comp[a[i]]) {
          ans--;
        }
        cnt[a[i]]++;
        if (cnt[a[i]] == comp[a[i]]) {
          ans++;
        }
      }
      for (int i = pl; i < blocks[b][q].l; i++) {
        if (cnt[a[i]] == comp[a[i]]) {
          ans--;
        }
        cnt[a[i]]--;
        if (cnt[a[i]] == comp[a[i]]) {
          ans++;
        }
      }
      pl = blocks[b][q].l;
      res[blocks[b][q].num] = ans;
    }
  }
  for (int i = 0; i < m; i++) {
    cout << res[i] << "\n";
  }
  return 0;
}
