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;
vector<int> v;
int a[N], ans[N], dp[480][N];
int main() {
  int n, m;
  cin >> n >> m;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    if (a[i] <= 100000 && a[i] >= 1) {
      ans[a[i]]++;
    }
  }
  for (int i = 1; i <= n; i++) {
    if (ans[i] >= i) {
      v.push_back(i);
    }
  }
  for (int i = 0; i < v.size(); i++) {
    dp[i][0] = 0;
    for (int j = 1; j <= n; j++) {
      if (a[j - 1] == v[i]) {
        dp[i][j] = dp[i][j - 1] + 1;
      } else {
        dp[i][j] = dp[i][j - 1];
      }
    }
  }
  while (m--) {
    long long l, r;
    cin >> l >> r;
    long long sum = 0;
    for (int i = 0; i < v.size(); i++) {
      if (dp[i][r] - dp[i][l - 1] == v[i]) {
        sum++;
      }
    }
    cout << sum << endl;
  }
  return 0;
}
