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 a[100010], cnt[100010], ans[100010], sum[100010], l[100010], r[100010];
int main() {
  int n, m;
  scanf("%d", &n);
  scanf("%d", &m);
  int x;
  for (int i = 0; i < n; i++) {
    scanf("%d", &x);
    a[i] = x;
    if (x <= n) cnt[a[i]]++;
  }
  int y, z;
  for (int i = 0; i < m; i++) {
    scanf("%d", &y), scanf("%d", &z);
    l[i] = y;
    r[i] = z;
  }
  for (int i = 1; i < n + 1; i++) {
    if (cnt[i] >= i) {
      sum[0] = 0;
      for (int j = 0; j < n; j++) sum[j + 1] = sum[j] + (a[j] == i ? 1 : 0);
      for (int j = 0; j < m; j++)
        if (sum[r[j]] - sum[l[j] - 1] == i) ans[j]++;
    }
  }
  for (int i = 0; i < m; i++) printf("%d\n", ans[i]);
}
