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 long long mx = 1e5 + 10;
long long n, m, a[mx], cnt[mx], r[mx], l[mx], p[mx], ans[mx];
int main() {
  cin >> n >> m;
  for (long long i = 1; i <= n; i++) {
    cin >> a[i];
    if (a[i] < mx) cnt[a[i]]++;
  }
  for (long long i = 0; i < m; i++) cin >> l[i] >> r[i];
  for (long long in = 1; in < mx; in++) {
    if (cnt[in] >= in) {
      p[0] = 0;
      for (long long i = 1; i <= n; i++) p[i] = p[i - 1] + (a[i] == in);
      for (long long i = 0; i < m; i++)
        ans[i] += ((p[r[i]] - p[l[i] - 1]) == in);
    }
  }
  for (long long i = 0; i < m; i++) cout << ans[i] << endl;
  return 0;
}
