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;
const int MOD = 1e9 + 7;
const int MAX = 1e5 + 5;
void solve() {
  int n, m;
  cin >> n >> m;
  vector<long long> arr(n + 1, 0);
  vector<int> cnt(n + 1, 0);
  for (int i = 1; i <= n; i++) {
    cin >> arr[i];
    if (arr[i] <= n) cnt[arr[i]]++;
  }
  pair<int, int> Q[m];
  for (int i = 0; i < m; i++) cin >> Q[i].first >> Q[i].second;
  vector<int> ans(m, 0);
  for (int i = 1; i <= n; i++) {
    if (cnt[i] < i) continue;
    vector<int> temp(n + 1, 0);
    for (int j = 1; j <= n; j++) {
      temp[j] = temp[j - 1] + (arr[j] == i);
    }
    for (int j = 0; j < m; j++) {
      int l = Q[j].first, r = Q[j].second;
      if (temp[r] - temp[l - 1] == i) ans[j]++;
    }
  }
  for (int i = 0; i < m; i++) cout << ans[i] << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  solve();
}
