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 main(void) {
  long n, m, l, r;
  vector<long> a(100005);
  map<long, vector<long> >::iterator mi;
  vector<long>::iterator vi1, vi2;
  map<long, vector<long> > hash;
  scanf("%ld %ld", &n, &m);
  for (long i = 0; i < n; i++) {
    scanf("%ld", &a[i]);
    if (hash.find(a[i]) != hash.end())
      hash[a[i]].push_back(i);
    else {
      vector<long> v;
      v.push_back(i);
      hash[a[i]] = v;
    }
  }
  mi = hash.begin();
  while (mi != hash.end()) {
    if (mi->first > mi->second.size()) {
      long temp = mi->first;
      mi++;
      hash.erase(hash.find(temp));
    } else
      mi++;
  }
  long diff = r - l + 1, count = 0;
  long first, last;
  for (long i = 0; i < m; i++) {
    scanf("%ld %ld", &l, &r);
    diff = r - l + 1, count = 0;
    mi = hash.begin();
    while (mi->first <= diff && mi != hash.end()) {
      vi1 = upper_bound(mi->second.begin(), mi->second.end(), r - 1);
      vi2 = lower_bound(mi->second.begin(), mi->second.end(), l - 1);
      last = vi1 - mi->second.begin(), first = vi2 - mi->second.begin();
      if (last - first == mi->first) count++;
      mi++;
    }
    printf("%ld\n", count);
  }
  return 0;
}
