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);
  map<int, int> freq;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    freq[a[i]]++;
  }
  vector<vector<int> > check;
  check.reserve(2 * sqrt(n) + 1);
  for (auto& p : freq) {
    if (p.second < p.first) continue;
    vector<int> f(n + 1);
    f[0] = p.first;
    for (int i = 1; i <= n; i++) {
      f[i] = (a[i - 1] == p.first);
      if (i > 1) f[i] += f[i - 1];
    }
    check.push_back(f);
  }
  while (m--) {
    int l, r;
    cin >> l >> r;
    int z = 0;
    for (auto& f : check) {
      int c = f[r] - (l > 1 ? f[l - 1] : 0);
      if (c == f[0]) z++;
    }
    cout << z << endl;
  }
  return 0;
}
