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 = 100009;
int sol, qs[N], qe[N], sorted[N], ans[N], a[N], s, e;
unordered_map<int, int> freq;
void add(int el) {
  sol -= !(freq[el] ^ el);
  freq[el]++;
  sol += !(freq[el] ^ el);
}
void remove(int el) {
  sol -= !(freq[el] ^ el);
  freq[el]--;
  sol += !(freq[el] ^ el);
}
void mo(int idx) {
  while (s > qs[idx]) add(a[--s]);
  while (e <= qe[idx]) add(a[e++]);
  while (s < qs[idx]) remove(a[s++]);
  while (e > qe[idx] + 1) remove(a[--e]);
  ans[idx] = sol;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n, m;
  cin >> n >> m;
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < m; i++) {
    cin >> qs[i] >> qe[i];
    qs[i]--, qe[i]--;
    sorted[i] = i;
  }
  int sq = sqrt(n);
  sort(sorted, sorted + m, [sq](int i, int j) {
    return make_pair(qs[i] / sq, qe[i]) < make_pair(qs[j] / sq, qe[j]);
  });
  for (int i = 0; i < m; i++) mo(sorted[i]);
  for (int i = 0; i < m; i++) cout << ans[i] << "\n";
}
