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 = 100005;
int cnt[N];
int sum[500][N];
int main(int argc, char *argv[]) {
  int n, m;
  cin >> n >> m;
  int a[n];
  for (int i = (0), _b = (n); i < (_b); ++i) cin >> a[i];
  for (int i = (0), _b = (n); i < (_b); ++i)
    if (a[i] <= n) cnt[a[i]]++;
  vector<int> v;
  for (int i = (1), _b = (n + 1); i < (_b); ++i)
    if (cnt[i] >= i) v.push_back(i);
  for (int i = (0), _b = (v.size()); i < (_b); ++i)
    for (int j = (0), _b = (n); j < (_b); ++j)
      sum[i][j + 1] = sum[i][j] + ((a[j] == v[i]) ? 1 : 0);
  int l, r;
  while (m--) {
    cin >> l >> r;
    --l;
    int ans = 0;
    for (int i = (0), _b = (v.size()); i < (_b); ++i) {
      if (sum[i][r] - sum[i][l] == v[i]) ++ans;
    }
    cout << ans << endl;
  }
  return 0;
}
