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;
int n, m;
int a[100010], cnt[100010];
int b[500][100010], w[500];
int cw = 0;
int main() {
  cin >> n >> m;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    if (a[i] <= 100000) ++cnt[a[i]];
  }
  for (int num = 1; num <= 100000; ++num)
    if (cnt[num] >= num) {
      if (a[0] == num) b[cw][0] = 1;
      w[cw] = num;
      for (int i = 1; i < n; i++)
        if (a[i] == num)
          b[cw][i] = b[cw][i - 1] + 1;
        else
          b[cw][i] = b[cw][i - 1];
      ++cw;
    }
  for (int i = 0; i < m; i++) {
    int ans = 0;
    int l, r;
    cin >> l >> r;
    --l;
    --r;
    for (int k = 0; k < cw; k++) {
      bool flag = false;
      if (l == 0) {
        if (b[k][r] == w[k]) flag = true;
      } else {
        if (b[k][r] - b[k][l - 1] == w[k]) flag = true;
      }
      if (flag) ++ans;
    }
    cout << ans << endl;
  }
}
