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::sync_with_stdio(false);
  int n, m;
  cin >> n >> m;
  vector<int> a(n), l(m), r(m), c(n + 1);
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
    if (a[i] <= n) {
      ++c[a[i]];
    }
  }
  for (int i = 0; i < m; ++i) {
    cin >> l[i] >> r[i];
    --l[i], --r[i];
  }
  vector<int> ans(m);
  for (int i = 1; i <= n; ++i) {
    if (c[i] >= i) {
      vector<int> num(n);
      num[0] = (a[0] == i ? 1 : 0);
      for (int j = 1; j < n; ++j)
        num[j] = (a[j] == i ? num[j - 1] + 1 : num[j - 1]);
      for (int j = 0; j < m; ++j) {
        int x = (l[j] == 0 ? num[r[j]] : num[r[j]] - num[l[j] - 1]);
        if (x == i) ++ans[j];
      }
    }
  }
  for (int i = 0; i < m; ++i) {
    cout << ans[i] << endl;
  }
  return 0;
}
