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 n, m;
int a[100010];
int c[100010];
int ans[100010];
vector<int> vv[100010];
struct node {
  int l, r;
  int id;
  bool operator<(node b) const { return r < b.r; }
} q[100010];
int lowbit(int x) { return x & -x; }
void update(int x, int val) {
  while (x <= n) {
    c[x] += val;
    x += lowbit(x);
  }
}
int sum(int x) {
  int ret = 0;
  while (x) {
    ret += c[x];
    x -= lowbit(x);
  }
  return ret;
}
int main() {
  cin >> n >> m;
  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);
  for (int i = 1, j = 1; i <= n; i++) {
    if (a[i] <= n) {
      vv[a[i]].push_back(i);
      int sz = vv[a[i]].size();
      if (sz >= a[i]) {
        update(vv[a[i]][sz - a[i]], 1);
        if (sz > a[i]) update(vv[a[i]][sz - a[i] - 1], -2);
        if (sz > a[i] + 1) update(vv[a[i]][sz - a[i] - 2], 1);
      }
    }
    while (i == q[j].r && j <= m) {
      ans[q[j].id] = sum(q[j].r) - sum(q[j].l - 1);
      j++;
    }
  }
  for (int i = 1; i <= m; i++) cout << ans[i] << endl;
  return 0;
}
