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;
vector<long long> occ[100001];
vector<long long> check;
int main() {
  std::ios::sync_with_stdio(false);
  long long n, m, a, b;
  scanf("%lld %lld", &n, &m);
  for (__typeof(n) i = (0) - ((0) > (n)); i != (n) - ((0) > (n));
       i += 1 - 2 * ((0) > (n))) {
    scanf("%lld", &a);
    if (a > 100000) continue;
    occ[a].push_back(i);
  }
  for (int i = 1; i <= 100000; i++) {
    if (occ[i].size() >= i) check.push_back(i);
  }
  long long k = check.size();
  for (__typeof(m) i = (0) - ((0) > (m)); i != (m) - ((0) > (m));
       i += 1 - 2 * ((0) > (m))) {
    scanf("%lld %lld", &a, &b);
    a--;
    b--;
    long long ans = 0;
    for (int j = 0; j < k; j++) {
      auto atual = check[j];
      auto& v = occ[atual];
      auto lb = lower_bound(v.begin(), v.end(), a);
      auto ub = upper_bound(v.begin(), v.end(), b);
      if (ub - lb == check[j]) ans++;
    }
    printf("%lld\n", ans);
  }
}
