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 = 1e5 + 2;
int main() {
  int n, m;
  scanf("%d %d", &n, &m);
  vector<int> a(n + 1), l(m), r(m), freq(n + 1, 0);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    if (a[i] <= n) {
      freq[a[i]]++;
    }
  }
  for (int i = 0; i < m; i++) {
    scanf("%d %d", &l[i], &r[i]);
  }
  vector<int> ans(m, 0), cand;
  for (int i = 1; i <= n; i++) {
    if (freq[i] >= i) cand.push_back(i);
  }
  for (int x : cand) {
    vector<int> cnt(n + 1, 0);
    for (int i = 1; i <= n; i++)
      cnt[i] += (a[i] == x ? 1 + cnt[i - 1] : cnt[i - 1]);
    int res = 0;
    for (int i = 0; i < m; i++) ans[i] += (cnt[r[i]] - cnt[(l[i] - 1)] == x);
  }
  for (int val : ans) printf("%d\n", val);
  return 0;
}
