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;
map<int, int> freq;
bool vis[100010];
map<int, int> m;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n, m;
  cin >> n >> m;
  int a[n];
  vector<int> v1, v;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    ++freq[a[i]];
    if (freq[a[i]] >= a[i]) v1.push_back(a[i]);
  }
  sort(v1.begin(), v1.end());
  int sz = unique(v1.begin(), v1.end()) - v1.begin();
  for (int i = 0; i < sz; i++) v.push_back(v1[i]);
  vector<int> vi[460];
  for (int i = 0; i < 460; i++) vi[i].resize(100005);
  for (int i = 0; i < v.size(); i++) {
    int l = 0;
    for (int j = 0; j < n; j++) {
      vi[i][j] += vi[i][max(j - 1, 0)] + (a[j] == v[i]);
      if (a[j] == v[i]) l = j;
    }
    vi[i].resize(l + 1);
    vi[i].resize(100005, vi[i][l]);
  }
  for (int i = 0; i < m; i++) {
    int l, r;
    cin >> l >> r;
    l--, r--;
    int cnt = 0;
    for (int i = 0; i < v.size(); i++) {
      if (l) {
        if (vi[i][r] - vi[i][l - 1] == v[i]) cnt++;
      } else {
        if (vi[i][r] == v[i]) cnt++;
      }
    }
    cout << cnt << endl;
  }
}
