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;
struct Cnt {
  int num;
  int count;
} cnt[501];
int len = 0;
int a[100005] = {0};
int tmp[100005] = {0};
int sum[501][100005] = {0};
int main() {
  int n, m;
  cin >> n >> m;
  for (int i = 1; i <= n; i++) cin >> a[i];
  memcpy(tmp, a, sizeof(tmp));
  sort(a + 1, a + n + 1);
  for (int i = 1; i <= n && a[i] <= n; i++) {
    int c = 1;
    while (a[i] == a[i + 1]) c++, i++;
    if (c < a[i]) continue;
    cnt[len++] = {a[i], c};
  }
  for (int i = 1; i <= n; i++) {
    bool find = false;
    int k = 0;
    for (int j = 0; j < len; j++) {
      if (cnt[j].num == tmp[i]) {
        find = true, k = j;
        break;
      }
    }
    if (find) sum[k][i]++;
    for (int j = 0; j < len; j++) {
      sum[j][i] += sum[j][i - 1];
    }
  }
  for (int i = 1; i <= m; i++) {
    int l, r;
    cin >> l >> r;
    int ans = 0;
    for (int j = 0; j < len; j++) {
      if (sum[j][r] - sum[j][l - 1] == cnt[j].num) ans++;
    }
    cout << ans << "\n";
  }
  return 0;
}
