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 tot[450][100005], a[100005], con[100005], inv[450];
int main() {
  int n, m, low, high, t = 0, cnt;
  memset(con, 0, sizeof(con));
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &a[i]);
    if (a[i] <= n) ++con[a[i]];
  }
  for (int i = 1; i <= n; ++i) {
    if (con[i] >= i) {
      inv[t] = i;
      tot[t++][0] = 0;
    }
  }
  for (int i = 1; i <= n; ++i) {
    for (int j = 0; j < t; ++j) {
      tot[j][i] = tot[j][i - 1];
      if (inv[j] == a[i]) ++tot[j][i];
    }
  }
  while (m--) {
    scanf("%d%d", &low, &high);
    cnt = 0;
    for (int i = 0; i < t; ++i) {
      if (tot[i][high] - tot[i][low - 1] == inv[i]) ++cnt;
    }
    cout << (cnt) << endl;
  }
  return 0;
}
