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 mapa[100002][500];
int tab[1000002], num[1000002], kub[1000002];
vector<int> v;
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n, q;
  cin >> n >> q;
  for (int i = 1; i <= n; i++) {
    cin >> tab[i];
    if (tab[i] <= 1000000) kub[tab[i]]++;
  }
  for (int i = 1; i <= n; i++) {
    if (kub[i] >= i) {
      v.push_back(i);
      num[i] = v.size() - 1;
    }
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < v.size(); j++) {
      mapa[i][j] = mapa[i - 1][j];
    }
    if (tab[i] <= 1000000 && kub[tab[i]] >= tab[i]) {
      mapa[i][num[tab[i]]]++;
    }
  }
  for (int ii = 0; ii < q; ii++) {
    int l, r, wyn = 0;
    cin >> l >> r;
    for (int j = 0; j < v.size(); j++) {
      if (mapa[r][j] - mapa[l - 1][j] == v[j]) wyn++;
    }
    cout << wyn << "\n";
  }
}
