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 gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long lb = 1000;
void vrikodara(long long n = 12) {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cout << setprecision(n) << fixed;
}
int32_t main() {
  vrikodara();
  long long n, m;
  cin >> n >> m;
  vector<long long> a(n + 1);
  vector<vector<long long>> t(n + 1);
  for (long long i = 1; i <= n; i++) {
    cin >> a[i];
    if (a[i] > n) continue;
    t[a[i]].push_back(i);
  }
  vector<vector<long long>> v;
  a.clear();
  for (long long i = 1; i <= n; i++) {
    if (t[i].size() >= i) {
      a.push_back(i);
      v.push_back(t[i]);
    }
  }
  long long idx = v.size();
  while (m--) {
    long long l, r;
    cin >> l >> r;
    long long d = r - l + 1;
    long long c = 0;
    for (long long i = 0; i < idx; i++) {
      long long val = a[i];
      if (val > d) break;
      long long sz = v[i].size();
      if (v[i][sz - 1] < l || v[i][0] > r) continue;
      long long lb = lower_bound(v[i].begin(), v[i].end(), l) - v[i].begin();
      long long rb = lower_bound(v[i].begin(), v[i].end(), r) - v[i].begin();
      if (lb >= sz) continue;
      if (rb != sz && v[i][rb] == r) rb++;
      if (rb - lb == val) {
        d -= val;
        c++;
      }
    }
    printf("%d\n", c);
  }
  return 0;
}
