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;
const int N = 1e6 + 5;
int a[N], root, cur, ans[N];
int cnt[N], n, m;
struct viet {
  int l, r, id;
};
viet q[N];
bool cmp(viet x, viet y) {
  if (x.l / root != y.l / root) return x.l / root < y.l / root;
  return x.r < y.r;
}
void add(int u) {
  if (a[u] > n) return;
  if (cnt[a[u]] == a[u]) cur--;
  cnt[a[u]]++;
  if (cnt[a[u]] == a[u]) cur++;
}
void del(int u) {
  if (a[u] > n) return;
  if (cnt[a[u]] == a[u]) cur--;
  cnt[a[u]]--;
  if (cnt[a[u]] == a[u]) cur++;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m;
  root = sqrt(n);
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 1; i <= m; i++) {
    cin >> q[i].l >> q[i].r;
    q[i].id = i;
  }
  sort(q + 1, q + m + 1, cmp);
  for (int i = q[1].l; i <= q[1].r; i++) {
    add(i);
  }
  int nowl = q[1].l, nowr = q[1].r;
  ans[q[1].id] = cur;
  for (int i = 2; i <= m; i++) {
    while (nowl < q[i].l) {
      del(nowl);
      nowl++;
    }
    while (nowl > q[i].l) {
      nowl--;
      add(nowl);
    }
    while (nowr > q[i].r) {
      del(nowr);
      nowr--;
    }
    while (nowr < q[i].r) {
      nowr++;
      add(nowr);
    }
    ans[q[i].id] = cur;
  }
  for (int i = 1; i <= m; i++) cout << ans[i] << '\n';
}
