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, M = 450;
int n, m;
int a[N], val[N], pref[M][N];
map<int, int> cnt;
int main() {
  ios_base ::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n >> m;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
    cnt[a[i]]++;
  }
  int sz = 0;
  for (auto it : cnt) {
    if (it.first > it.second) continue;
    ++sz;
    val[sz] = it.first;
    for (int i = 1; i <= n; ++i) {
      pref[sz][i] = pref[sz][i - 1];
      if (a[i] == val[sz]) pref[sz][i]++;
    }
  }
  for (int i = 1; i <= m; ++i) {
    int l, r;
    cin >> l >> r;
    int res = 0;
    for (int j = 1; j <= sz; ++j) {
      if (pref[j][r] - pref[j][l - 1] == val[j]) ++res;
    }
    cout << res << "\n";
  }
  return 0;
}
