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 = 100065;
int a[maxn];
int n, m;
int sum[maxn][500];
vector<int> v;
map<int, int> mp;
int cnt[maxn];
int main() {
  scanf("%d%d", &n, &m);
  mp.clear();
  v.clear();
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
  }
  memset(cnt, 0, sizeof(cnt));
  for (int i = 1; i <= n; i++) {
    if (a[i] > n) continue;
    cnt[a[i]]++;
  }
  int count = 0;
  for (int i = 1; i <= n; i++) {
    if (cnt[i] >= i) {
      mp[i] = count++;
      v.push_back(i);
    }
  }
  memset(sum, 0, sizeof(sum));
  for (int i = 1; i <= n; i++) {
    map<int, int>::iterator it;
    it = mp.find(a[i]);
    memcpy(sum[i], sum[i - 1], sizeof(sum[i]));
    if (it != mp.end()) {
      sum[i][it->second]++;
    }
  }
  while (m--) {
    int l, r;
    int ans = 0;
    scanf("%d%d", &l, &r);
    for (int i = 0; i < v.size(); i++) {
      int tmp = sum[r][i] - sum[l - 1][i];
      if (tmp == v[i]) ans++;
    }
    cout << ans << endl;
  }
}
