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;
long long n, m, a[100005];
long long M[100005];
struct que {
  long long ind;
  long long left;
  long long right;
} que[100005];
long long ans[100005];
long long block;
bool comp(struct que &q1, struct que &q2) {
  if (q1.left / block != q2.left / block) {
    return q1.left < q2.left;
  } else {
    return q1.right < q2.right;
  }
}
int cnt;
inline void add(long long x) {
  if (x > n) return;
  if (M[x] == x) cnt--;
  M[x]++;
  if (M[x] == x) cnt++;
}
inline void remove(long long x) {
  if (x > n) return;
  if (M[x] == x) cnt--;
  M[x]--;
  if (M[x] == x) cnt++;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  long long i, j, k;
  cin >> n >> m;
  block = sqrt(n);
  for (i = 0; i < n; i++) cin >> a[i];
  for (i = 0; i < m; i++) {
    cin >> que[i].left >> que[i].right;
    que[i].ind = i;
  }
  sort(que, que + m, comp);
  long long mo_left = 0, mo_right = -1;
  for (i = 0; i < m; i++) {
    long long left = que[i].left;
    long long right = que[i].right;
    left--, right--;
    while (mo_right < right) {
      mo_right++;
      add(a[mo_right]);
    }
    while (mo_right > right) {
      remove(a[mo_right]);
      mo_right--;
    }
    while (mo_left < left) {
      remove(a[mo_left]);
      mo_left++;
    }
    while (mo_left > left) {
      mo_left--;
      add(a[mo_left]);
    }
    ans[que[i].ind] = cnt;
  }
  for (i = 0; i < m; i++) {
    cout << ans[i] << endl;
  }
  return 0;
}
