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;
vector<int> num[100000];
struct query {
  int l, r;
  int id;
  bool operator<(const query &x) const { return l < x.l; }
} qur[100000];
int ans[100000], a[100000], val[100000];
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  map<int, int> mp;
  int L = 0;
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    if (mp.find(a[i]) == mp.end()) {
      val[L] = a[i];
      mp[a[i]] = L++;
    }
    num[mp[a[i]]].push_back(i);
  }
  for (int i = 0; i < m; i++) {
    scanf("%d%d", &qur[i].l, &qur[i].r);
    qur[i].l--;
    qur[i].r--;
    qur[i].id = i;
  }
  sort(qur, qur + m);
  for (int i = 0; i < n; i++)
    if (num[i].size() >= val[i])
      for (int j = 0, k = 0; j < num[i].size() && k < m; j++)
        while (k < m && qur[k].l <= num[i][j]) {
          if (j + val[i] - 1 < num[i].size() &&
              qur[k].r >= num[i][j + val[i] - 1] &&
              (j + val[i] == num[i].size() || qur[k].r < num[i][j + val[i]]))
            ans[qur[k].id]++;
          k++;
        }
  for (int i = 0; i < m; i++) printf("%d\n", ans[i]);
  return 0;
}
