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 MAXN = 100000 + 10;
int n, que;
int a[MAXN], cnt[MAXN], lo[MAXN], hi[MAXN], ans[MAXN];
map<int, int> mp;
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n >> que;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    if (mp.find(a[i]) == mp.end()) mp[a[i]] = 0;
    mp[a[i]]++;
  }
  vector<int> store;
  for (map<int, int>::iterator it = mp.begin(); it != mp.end(); it++)
    if ((it->second) >= (it->first)) store.push_back(it->first);
  for (int i = 1; i <= que; i++) cin >> lo[i] >> hi[i];
  for (int rep = 0; rep < (int)store.size(); rep++) {
    cnt[0] = 0;
    for (int i = 1; i <= n; i++) cnt[i] = cnt[i - 1] + (a[i] == store[rep]);
    for (int i = 1; i <= que; i++)
      if ((cnt[hi[i]] - cnt[lo[i] - 1]) == store[rep]) ans[i]++;
  }
  for (int i = 1; i <= que; i++) cout << ans[i] << endl;
  return 0;
}
