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, i, j, k, w, l, r, cnt, a[100100], c[100100], o[100100],
    d[100100][452];
vector<int> b;
int main() {
  scanf("%d%d", &n, &m);
  for (i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    if (a[i] <= n) c[a[i]]++;
  }
  for (i = 1; i <= n; i++)
    if (c[i] >= i) {
      b.push_back(i);
      o[i] = b.size();
    }
  k = b.size();
  for (i = 1; i <= n; i++) {
    for (j = 0; j < k; j++) d[i][j] = d[i - 1][j];
    if (a[i] <= n) {
      w = o[a[i]] - 1;
      if (w >= 0) d[i][w]++;
    }
  }
  while (m--) {
    scanf("%d%d", &l, &r);
    l--;
    for (cnt = i = 0; i < k; i++)
      if (d[r][i] - d[l][i] == b[i]) cnt++;
    printf("%d\n", cnt);
  }
  return 0;
}
