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;
void solve() {}
const long long maxn = 100500;
long long a[maxn], cnt[maxn], psa[maxn];
long long l[maxn], r[maxn], ans[maxn];
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  long long n, m;
  cin >> n >> m;
  for (long long i = 1; i <= n; i++) {
    cin >> a[i];
    if (a[i] <= 1e5) {
      cnt[a[i]]++;
    }
  }
  vector<long long> valid;
  for (long long i = 1; i <= 1e5; i++) {
    if (cnt[i] >= i) valid.emplace_back(i);
  }
  for (long long i = 1; i <= m; i++) {
    cin >> l[i] >> r[i];
  }
  for (long long x : valid) {
    for (long long i = 0; i < maxn; i++) {
      psa[i] = 0;
    }
    for (long long i = 1; i <= n; i++) {
      if (a[i] == x) psa[i]++;
      psa[i] += psa[i - 1];
    }
    for (long long i = 1; i <= m; i++) {
      if (psa[r[i]] - psa[l[i] - 1] == x) {
        ans[i]++;
      }
    }
  }
  for (long long i = 1; i <= m; i++) {
    cout << ans[i] << '\n';
  }
}
