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 a[100009];
struct query {
  int l, r, id;
  bool operator<(const query& b) const { return r < b.r; }
} q[100009];
int n, m;
int ans[100009];
vector<int> pos[100009];
int c[100009];
int sum(int i) {
  int ans = 0;
  while (i > 0) {
    ans += c[i];
    i -= (i & (-i));
  }
  return ans;
}
void update(int i, int x) {
  while (i <= n) {
    c[i] += x;
    i += (i & -i);
  }
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) {
    scanf("%d", a + i);
  }
  for (int i = 1; i <= m; i++) {
    scanf("%d %d", &q[i].l, &q[i].r);
    q[i].id = i;
  }
  sort(q + 1, q + 1 + m);
  for (int i = 1, j = 1; i <= n && j <= m; i++) {
    if (a[i] <= n) {
      vector<int>& v = pos[a[i]];
      v.push_back(i);
      if (v.size() >= a[i]) update(v[v.size() - a[i]], 1);
      if (v.size() > a[i]) update(v[v.size() - a[i] - 1], -2);
      if (v.size() > a[i] + 1) update(v[v.size() - a[i] - 2], 1);
    }
    while (q[j].r <= i && j <= m) {
      ans[q[j].id] = sum(q[j].r) - sum(q[j].l - 1);
      j++;
    }
  }
  for (int i = 1; i <= m; i++) {
    printf("%d\n", ans[i]);
  }
  return 0;
}
