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>
int N, Q, Am[450][100005], A[100005];
using namespace std;
vector<int> O[100005], U;
int main() {
  cin >> N >> Q;
  for (int i = 1; i <= N; i++) {
    cin >> A[i];
    if (A[i] <= N) O[A[i]].push_back(i);
  }
  for (int i = 1; i <= N; i++)
    if (O[i].size() >= i) U.push_back(i);
  for (int i = 0; i < U.size(); i++)
    for (int p = 1; p <= N; p++) Am[i][p] = Am[i][p - 1] + (A[p] == U[i]);
  while (Q--) {
    int l, r;
    cin >> l >> r;
    int ans = 0;
    for (int i = 0; i < U.size(); i++)
      if (Am[i][r] - Am[i][l - 1] == U[i]) ans++;
    cout << ans << "\n";
  }
}
