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;
int n, m, x, l, r;
vector<int> v[111111];
vector<int> cand;
int ans;
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 0; i < n; i++) {
    scanf("%d", &x);
    if (x <= n) v[x].push_back(i + 1);
  }
  for (int i = 1; i <= n; i++)
    if (v[i].size() >= i) cand.push_back(i);
  for (int i = 0; i < m; i++) {
    scanf("%d%d", &l, &r);
    ans = 0;
    for (int j = 0; j < cand.size(); j++) {
      x = cand[j];
      int cnt = upper_bound(v[x].begin(), v[x].end(), r) -
                lower_bound(v[x].begin(), v[x].end(), l);
      if (cnt == x) ans++;
    }
    printf("%d\n", ans);
  }
  return 0;
}
