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 N = 100005;
int ans[N], cas[N], num[N], a[N][2], tmp[N], sum[N];
int main() {
  int n, m;
  while (scanf("%d%d", &n, &m) != EOF) {
    memset(tmp, 0, sizeof(tmp));
    memset(sum, 0, sizeof(sum));
    memset(cas, 0, sizeof(cas));
    for (int i = 1; i <= n; i++) {
      scanf("%d", &num[i]);
      if (num[i] <= n) tmp[num[i]]++;
    }
    for (int i = 1; i <= m; i++) {
      scanf("%d%d", &a[i][0], &a[i][1]);
    }
    int len = 0;
    for (int i = 1; i <= n; i++) {
      if (tmp[i] >= i) ans[len++] = i;
    }
    for (int i = 0; i < len; i++) {
      int val = ans[i];
      sum[0] = 0;
      sum[1] = (val == num[1]);
      for (int j = 2; j <= n; j++) {
        sum[j] = sum[j - 1] + (num[j] == val);
      }
      for (int j = 1; j <= m; j++) {
        int x = a[j][0], y = a[j][1];
        cas[j] += (val == sum[y] - sum[x - 1]);
      }
    }
    for (int i = 1; i <= m; i++) {
      printf("%d\n", cas[i]);
    }
  }
  return 0;
}
