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, a[100010], c[100010];
int key[500], numKey = 0, sum[500][100010];
int main() {
  cin >> n >> m;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    c[i] = a[i];
  }
  sort(c, c + n);
  c[n] = 0;
  int counter = 1;
  for (int i = 1; i < n + 1; i++) {
    if (c[i] != c[i - 1]) {
      if (counter >= c[i - 1]) key[numKey++] = c[i - 1];
      counter = 1;
    } else
      counter++;
  }
  for (int i = 0; i < numKey; i++) {
    if (a[0] == key[i])
      sum[i][0] = 1;
    else
      sum[i][0] = 0;
    for (int j = 1; j < n; j++) {
      if (a[j] == key[i])
        sum[i][j] = sum[i][j - 1] + 1;
      else
        sum[i][j] = sum[i][j - 1];
    }
  }
  for (int i = 0; i < m; i++) {
    int s, t, ans = 0;
    cin >> s >> t;
    s--;
    t--;
    for (int j = 0; j < numKey; j++) {
      if (s >= 1 && (sum[j][t] - sum[j][s - 1]) == key[j])
        ans++;
      else if (s == 0 && sum[j][t] == key[j])
        ans++;
    }
    cout << ans << endl;
  }
  return 0;
}
