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 d[501][100001];
int main() {
  int i, j, n, m, a[100001], l, r;
  map<int, int> num;
  int pos[501], le = 0;
  cin >> n >> m;
  for (i = 1; i <= n; i++) {
    cin >> a[i];
    num[a[i]]++;
  }
  for (map<int, int>::iterator it = num.begin(); it != num.end(); it++) {
    if (it->second >= it->first) pos[++le] = it->first;
  }
  for (i = 1; i < le; i++) d[i][0] = 0;
  for (i = 1; i <= le; i++)
    for (j = 1; j <= n; j++)
      if (pos[i] == a[j])
        d[i][j] = d[i][j - 1] + 1;
      else
        d[i][j] = d[i][j - 1];
  for (i = 1; i <= m; i++) {
    cin >> l >> r;
    int ans = 0;
    for (j = 1; j <= le; j++) {
      if (d[j][r] - d[j][l - 1] == pos[j]) ans++;
    }
    cout << ans << endl;
  }
  return 0;
}
