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() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t = 1;
  while (t--) {
    int n = 0, m = 0, k = 0, i = 0, j = 0, p = 0, q = 0, x = 0, y = 0, z = 0,
        ans = 0, cnt = 0;
    string s;
    bool flag = false;
    cin >> n >> m;
    int a[n];
    map<int, int> frq;
    for (i = 0; i < n; i++) {
      cin >> a[i];
      if (frq.find(a[i]) == frq.end())
        frq[a[i]] = 1;
      else
        frq[a[i]]++;
    }
    vector<int> v;
    for (auto u : frq) {
      if (u.second >= u.first) v.push_back(u.first);
    }
    sort(v.begin(), v.end());
    k = v.size();
    int dp[k][n + 1];
    memset(dp, 0, sizeof(dp));
    for (j = 1; j <= n; j++) {
      for (i = 0; i < k; i++) dp[i][j] = dp[i][j - 1];
      for (i = 0; i < k; i++) {
        if (a[j - 1] == v[i]) {
          dp[i][j]++;
          break;
        }
      }
    }
    while (m--) {
      cnt = 0;
      cin >> p >> q;
      for (i = 0; i < k; i++) {
        y = dp[i][q] - dp[i][p - 1];
        if (y == v[i]) cnt++;
      }
      cout << cnt << endl;
    }
  }
  return 0;
}
