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 n, m, l[100001], r[100001], ans[100001], a[100001], s[100001], b[100001];
int main() {
  cin >> m >> n;
  for (int i = 1; i <= m; i++) {
    cin >> a[i];
    if (a[i] <= m) b[a[i]]++;
  }
  for (int i = 1; i <= n; i++) {
    cin >> l[i] >> r[i];
  }
  for (int i = 1; i <= m; i++) {
    if (b[i] >= i) {
      for (int j = 1; j <= m; j++) {
        s[j] = s[j - 1] + (a[j] == i);
      }
      for (int j = 1; j <= n; j++)
        if (s[r[j]] - s[l[j] - 1] == i) ans[j]++;
    }
  }
  for (int i = 1; i <= n; i++) cout << ans[i] << endl;
  return 0;
}
