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 dp[100011][511];
int main() {
  int n, m;
  cin >> n >> m;
  vector<int> g(n + 1);
  map<int, int> dict;
  for (int i = 0; i < n; i++) {
    cin >> g[i];
    dict[g[i]]++;
  }
  vector<int> p;
  map<int, int>::iterator it = dict.begin();
  for (; it != dict.end(); it++)
    if (it->first <= it->second) p.push_back(it->first);
  for (int i = 0; i <= n; i++)
    for (int j = 0; j < p.size(); j++) {
      if (!i)
        dp[i][j] = 0;
      else {
        dp[i][j] = dp[i - 1][j];
        if (g[i - 1] == p[j]) dp[i][j]++;
      }
    }
  for (int i = 0; i < m; i++) {
    int l, r;
    cin >> l >> r;
    int ans = 0;
    for (int j = 0; j < p.size(); j++)
      if (dp[r][j] - dp[l - 1][j] == p[j]) ans++;
    cout << ans << endl;
  }
  return 0;
}
