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;
struct query {
  long long l, r, id;
};
long long n, a[(long long)2e5 + 10], sq, total_query;
bool cmp(struct query x, struct query y) {
  if (x.l / sq == y.l / sq) return x.r < y.r;
  return x.l / sq < y.l / sq;
}
long long occ[(long long)1e5 + 1] = {0};
long long cur_ans = 0;
map<int, int> m;
void add(long long x) {
  if (x > 1e5) return;
  occ[x]++;
  if (occ[x] == x)
    cur_ans++;
  else if (occ[x] - x == 1)
    cur_ans--;
}
void rem(long long x) {
  if (x > 1e5) return;
  occ[x]--;
  if (occ[x] == x)
    cur_ans++;
  else if (occ[x] - x == -1)
    cur_ans--;
}
int main(void) {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> total_query;
  sq = sqrt(n);
  for (__typeof(n) i = (0) - ((0) > (n)); i != (n) - ((0) > (n));
       i += 1 - 2 * ((0) > (n)))
    cin >> a[i];
  struct query q[total_query];
  for (__typeof(total_query) i = (0) - ((0) > (total_query));
       i != (total_query) - ((0) > (total_query));
       i += 1 - 2 * ((0) > (total_query))) {
    cin >> q[i].l >> q[i].r;
    q[i].l--;
    q[i].r--;
    q[i].id = i;
  }
  sort(q, q + total_query, cmp);
  long long i = 0, j = -1;
  long long ans[total_query];
  for (auto p : q) {
    while (j < p.r) add(a[++j]);
    while (j > p.r) rem(a[j--]);
    while (i < p.l) rem(a[i++]);
    while (i > p.l) add(a[--i]);
    ans[p.id] = cur_ans;
  }
  for (long long i : ans) cout << i << '\n';
}
