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() {
  std::ios::sync_with_stdio(false);
  long long int n, m;
  cin >> n >> m;
  vector<long long int> arr(n);
  for (long long int i = 0; i < n; ++i) cin >> arr[i];
  vector<pair<long long int, long long int>> q;
  for (long long int i = 0; i < m; ++i) {
    long long int a, b;
    cin >> a >> b;
    q.push_back(make_pair(a, b));
  }
  long long int count[n + 1];
  for (long long int i = 0; i < n + 1; ++i) count[i] = 0;
  for (long long int i = 0; i < n; ++i) {
    if (arr[i] <= n) count[arr[i]]++;
  }
  long long int ans[m];
  for (long long int i = 0; i < m; ++i) ans[i] = 0;
  for (long long int i = 1; i <= n; i++) {
    if (count[i] >= i) {
      long long int cumsum[n + 1];
      cumsum[0] = 0;
      for (long long int j = 0; j < n; ++j) {
        if (j == 0)
          cumsum[j + 1] = (arr[j] == i) ? 1 : 0;
        else {
          cumsum[j + 1] = cumsum[j] + ((arr[j] == i) ? 1 : 0);
        }
      }
      for (long long int j = 0; j < m; ++j) {
        if (cumsum[q[j].second] - cumsum[q[j].first - 1] == i) ans[j]++;
      }
    }
  }
  for (long long int i = 0; i < m; ++i) cout << ans[i] << "\n";
  return 0;
}
