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), cnt(n + 1);
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
    if (a[i] <= n) ++cnt[a[i]];
  }
  vector<int> good;
  for (int i = 1; i < n + 1; ++i)
    if (cnt[i] >= i) good.push_back(i);
  int k = good.size();
  vector<vector<int> > pref(n + 1, vector<int>(k));
  for (int i = 0; i < n; ++i)
    for (int j = 0; j < k; ++j) pref[i + 1][j] = pref[i][j] + (a[i] == good[j]);
  while (m--) {
    int l, r;
    cin >> l >> r;
    int res = 0;
    for (int i = 0; i < k; ++i)
      if (pref[r][i] - pref[l - 1][i] == good[i]) ++res;
    cout << res << endl;
  }
  return 0;
}
