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() {
  int n, m;
  cin >> n >> m;
  vector<int> a(n), contVec(n + 1, 0);
  for (int ii = int(0); ii != int(n); ii++) {
    cin >> a[ii];
    if (a[ii] <= n) contVec[a[ii]]++;
  }
  vector<pair<int, int> > queries(m);
  for (int ii = int(0); ii != int(m); ii++) {
    int l, r;
    cin >> l >> r;
    queries[ii] = make_pair(l, r);
  }
  vector<int> ans(m, 0);
  for (int x = int(1); x != int(n + 1); x++) {
    if (contVec[x] >= x) {
      vector<int> sum(n + 1, 0);
      for (int ii = int(1); ii != int(n + 1); ii++)
        sum[ii] += sum[ii - 1] + ((x == a[ii - 1]) ? 1 : 0);
      for (int ii = int(0); ii != int(m); ii++) {
        if (sum[queries[ii].second] - sum[queries[ii].first - 1] == x)
          ans[ii]++;
      }
    }
  }
  for (int ii = int(0); ii != int(m); ii++) cout << ans[ii] << endl;
  return 0;
}
