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 + 1;
const int K = 666;
static int cnt[N], ans, _ans[N];
int main() {
  int n, q;
  cin >> n >> q;
  int a[n];
  for (int i = 0; i < n; ++i) cin >> a[i];
  int l[q], r[q], mo_l = 0, mo_r = 0;
  for (int i = 0; i < q; ++i) cin >> l[i] >> r[i], --l[i];
  int cnt_bl = n / K + bool(n % K);
  vector<int> block[cnt_bl];
  for (int i = 0; i < q; ++i) block[l[i] / K].push_back(i);
  for (int i = 0; i < cnt_bl; ++i) {
    sort(block[i].begin(), block[i].end(),
         [&](int a, int b) { return r[a] < r[b]; });
    for (int x : block[i]) {
      while (mo_l < l[x]) {
        if (a[mo_l] < N)
          ans -= cnt[a[mo_l]] == a[mo_l], ans += --cnt[a[mo_l]] == a[mo_l];
        mo_l++;
      }
      while (mo_l > l[x])
        if (a[--mo_l] < N)
          ans -= cnt[a[mo_l]] == a[mo_l], ans += ++cnt[a[mo_l]] == a[mo_l];
      while (mo_r < r[x]) {
        if (a[mo_r] < N)
          ans -= cnt[a[mo_r]] == a[mo_r], ans += ++cnt[a[mo_r]] == a[mo_r];
        mo_r++;
      }
      while (mo_r > r[x])
        if (a[--mo_r] < N)
          ans -= cnt[a[mo_r]] == a[mo_r], ans += --cnt[a[mo_r]] == a[mo_r];
      _ans[x] = ans;
    }
  }
  for (int i = 0; i < q; ++i) cout << _ans[i] << '\n';
  return 0;
}
