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;
const int N = (int)(5e5) + 322;
const long long inf = (long long)(1e18) + 123;
const int mod = 31011;
const double EPS = 1e-9;
int n, m, a[N], sz = 400, ans, l, r, res[N];
int cnt[N];
struct seg {
  int l, r, id;
} p[N];
bool cmp(seg a, seg b) {
  if (a.r / sz != b.r / sz) return a.r / sz < b.r / sz;
  return a.l < b.l;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n >> m;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
    if (a[i] > n) a[i] = N - 10;
  }
  for (int i = 1; i <= m; ++i) {
    cin >> p[i].l >> p[i].r;
    p[i].id = i;
  }
  a[0] = N - 10;
  sort(p + 1, p + m + 1, cmp);
  for (int i = 1; i <= m; ++i) {
    while (l < p[i].l) {
      if (a[l] == cnt[a[l]]) ans--;
      cnt[a[l]]--;
      if (a[l] == cnt[a[l]]) ans++;
      l++;
    }
    while (l > p[i].l) {
      l--;
      if (a[l] == cnt[a[l]]) ans--;
      cnt[a[l]]++;
      if (a[l] == cnt[a[l]]) ans++;
    }
    while (r < p[i].r) {
      r++;
      if (a[r] == cnt[a[r]]) ans--;
      cnt[a[r]]++;
      if (a[r] == cnt[a[r]]) ans++;
    }
    while (r > p[i].r) {
      if (a[r] == cnt[a[r]]) ans--;
      cnt[a[r]]--;
      if (a[r] == cnt[a[r]]) ans++;
      r--;
    }
    res[p[i].id] = ans;
  }
  for (int i = 1; i <= m; ++i) {
    cout << res[i] << "\n";
  }
  return 0;
}
