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;
long long a[100005];
map<long long, long long> mp;
bool seen[100005];
int kandid[2000];
int tedad[500][100005];
int main() {
  long long n, q;
  cin >> n >> q;
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < n; i++) {
    if (a[i] <= 1e5) mp[a[i]]++;
  }
  int k = 0;
  for (int i = 0; i < n; i++) {
    if (a[i] <= 1e5 and mp[a[i]] >= a[i] and !seen[a[i]]) {
      kandid[k] = a[i];
      seen[a[i]] = true;
      k++;
    }
  }
  sort(kandid, kandid + k);
  for (int i = 0; i < k; i++) {
    if (a[0] == kandid[i]) tedad[i][0]++;
  }
  for (int i = 1; i < n; i++) {
    for (int j = 0; j < k; j++) {
      tedad[j][i] = tedad[j][i - 1];
    }
    for (int j = 0; j < k; j++) {
      if (a[i] == kandid[j]) tedad[j][i]++;
    }
  }
  while (q--) {
    int l, r;
    cin >> l >> r;
    int ans = 0;
    if (r == 1) {
      if (a[0] == 1) ans++;
    } else {
      for (int i = 0; i < k; i++) {
        if (tedad[i][r - 1] - tedad[i][l - 2] == kandid[i]) ans++;
      }
    }
    cout << ans << endl;
  }
  return 0;
}
