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, q, num[100010];
pair<int, pair<int, int> > qry[100010];
int ans[100010];
vector<int> lst[100010];
int tree[100010];
void add(int pos, int v) {
  for (; pos <= n; pos += pos & (-pos)) tree[pos] += v;
}
int query(int pos) {
  int ret = 0;
  for (; pos; pos -= pos & (-pos)) ret += tree[pos];
  return ret;
}
int main() {
  int i, j, siz;
  scanf("%d%d", &n, &q);
  for (i = 1; i <= n; i++) scanf("%d", &num[i]);
  for (i = 1; i <= q; i++) {
    scanf("%d%d", &qry[i].second.first, &qry[i].first);
    qry[i].second.second = i;
  }
  sort(qry + 1, qry + 1 + q);
  for (i = j = 1; i <= n; i++)
    if (num[i] <= n) {
      lst[num[i]].push_back(i);
      if ((siz = lst[num[i]].size()) >= num[i]) {
        if (siz > num[i] + 1) add(lst[num[i]][siz - num[i] - 2], 1);
        if (siz > num[i]) add(lst[num[i]][siz - num[i] - 1], -2);
        add(lst[num[i]][siz - num[i]], 1);
      }
      for (; qry[j].first < i && j <= q; j++)
        ;
      for (; qry[j].first == i; j++)
        ans[qry[j].second.second] =
            query(qry[j].first) - query(qry[j].second.first - 1);
    }
  for (i = 1; i <= q; i++) printf("%d\n", ans[i]);
  return 0;
}
