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 + 10;
const int MOD = 1e9 + 7;
int a[N], cnt[N], c[505][N];
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n, q;
  cin >> n >> q;
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < n; i++)
    if (a[i] <= N) cnt[a[i]]++;
  vector<int> x;
  for (int i = 1; i < N; i++) {
    if (cnt[i] >= i) {
      x.push_back(i);
    }
  }
  for (int i = 0; i < x.size(); i++)
    for (int j = 0; j < N; j++) c[i][j + 1] = c[i][j] + (a[j] == x[i]);
  int l, r;
  while (q--) {
    cin >> l >> r;
    l--;
    int ans = 0;
    for (int i = 0; i < x.size(); i++)
      if (c[i][r] - c[i][l] == x[i]) ans++;
    cout << ans << '\n';
  }
  return 0;
}
