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 MAX = 1e5 + 2;
const long long MOD = 998244353;
vector<int> id[MAX];
int cnt[MAX];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n, m;
  cin >> n >> m;
  for (int i = 1; i <= n; i++) {
    int x;
    cin >> x;
    if (x < MAX) {
      id[x].push_back(i);
      cnt[x]++;
    }
  }
  vector<int> v;
  for (int i = 1; i < MAX; i++) {
    if (cnt[i] >= i) v.push_back(i);
  }
  while (m--) {
    int l, r, res = 0;
    cin >> l >> r;
    for (int x : v) {
      int hi = upper_bound(id[x].begin(), id[x].end(), r) - id[x].begin();
      int lo = upper_bound(id[x].begin(), id[x].end(), l - 1) - id[x].begin();
      if (hi - lo == x) res++;
    }
    cout << res << "\n";
  }
}
