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 int inf = 0x3f3f3f3f;
const int maxn = 100000 + 10;
const double pi = acos(-1.0);
int a[maxn], b[maxn], l[maxn], r[maxn];
int sum[maxn], ans[maxn];
int n, m;
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    if (a[i] <= n) b[a[i]]++;
  }
  for (int i = 1; i <= m; i++) {
    scanf("%d %d", &l[i], &r[i]);
  }
  for (int i = 1; i <= n; i++) {
    if (i <= b[i]) {
      for (int j = 1; j <= n; j++) {
        sum[j] = sum[j - 1] + (a[j] == i);
      }
      for (int j = 1; j <= m; j++) {
        int t = sum[r[j]] - sum[l[j] - 1];
        if (t == i) ans[j]++;
      }
    }
  }
  for (int i = 1; i <= m; i++) {
    printf("%d\n", ans[i]);
  }
  return 0;
}
