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 cnt[450][100005], b[100005];
vector<int> l;
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int TESTS = 1;
  while (TESTS--) {
    int n, q;
    cin >> n >> q;
    vector<int> a(n + 1);
    for (int i = 1; i < n + 1; i++) cin >> a[i];
    for (auto i : a)
      if (i <= n) b[i]++;
    for (int i = 1; i < n + 1; i++)
      if (b[i] >= i) l.push_back(i);
    for (int i = 0; i < (int)l.size(); i++) {
      int cur = l[i];
      for (int j = 1; j < n + 1; j++) {
        cnt[i][j] = cnt[i][j - 1];
        if (a[j] == cur) cnt[i][j]++;
      }
    }
    while (q--) {
      int x, y, ans = 0;
      cin >> x >> y;
      for (int j = 0; j < (int)l.size(); j++)
        if (cnt[j][y] - cnt[j][x - 1] == l[j]) ans++;
      cout << ans << '\n';
    }
  }
  return 0;
}
