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 c[200000];
int a[200000];
int b[200000], ans[200000];
int l[200000], r[200000];
int n, m, i, j;
int main() {
  cin >> n >> m;
  for (i = 1; i <= n; i++) cin >> a[i];
  for (i = 1; i <= n; i++)
    if (a[i] <= n) c[a[i]]++;
  for (i = 1; i <= m; i++) cin >> l[i] >> r[i];
  for (i = 1; i <= n; i++)
    if (c[i] >= i) {
      for (j = 1; j <= n; j++)
        if (a[j] == i)
          b[j] = b[j - 1] + 1;
        else
          b[j] = b[j - 1];
      for (j = 1; j <= m; j++)
        if (b[r[j]] - b[l[j] - 1] == i) ans[j]++;
    }
  for (j = 1; j <= m; j++) cout << ans[j] << endl;
}
