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 a[100001], b[100000], map[100000], c[634][100001];
int main() {
  int n, q;
  std::cin >> n >> q;
  for (int i = 1; i <= n; ++i) std::cin >> a[i];
  memcpy(b, &a[1], sizeof(b));
  std::stable_sort(b, b + n);
  int N = 0, i = 0;
  while (i < n) {
    int r = std::upper_bound(b, b + n, b[i]) - b;
    if (r - i >= b[i]) map[N++] = b[i];
    i = r;
  }
  for (int i = 0; i < N; ++i)
    for (int j = 1; j <= n; ++j) c[i][j] = c[i][j - 1] + (a[j] == map[i]);
  while (q--) {
    int l, r, count = 0;
    std::cin >> l >> r;
    for (int i = 0; i < N; ++i) count += map[i] == c[i][r] - c[i][l - 1];
    std::cout << count << '\n';
  }
  return 0;
}
