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 MxN = 1e5 + 5;
pair<int, int> q[MxN];
map<int, int> mp;
vector<int> vec;
int arr[MxN], ans[MxN];
int n, m;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m;
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
    mp[arr[i]]++;
  }
  for (int i = 1; i <= m; i++) cin >> q[i].first >> q[i].second;
  for (auto p : mp)
    if (p.second >= p.first) vec.push_back(p.first);
  for (auto v : vec) {
    vector<int> pref(n + 1, 0);
    for (int i = 1; i <= n; i++) pref[i] = pref[i - 1] + (arr[i - 1] == v);
    for (int i = 1; i <= m; i++)
      ans[i] += (pref[q[i].second] - pref[q[i].first - 1] == v);
  }
  for (int i = 1; i <= m; i++) cout << ans[i] << "\n";
  return 0;
}
