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 long long maxn = 1e5 + 5;
const long long sq = 317;
unordered_map<long long, long long> ab, ba;
long long n, m, res, arr[maxn], ans[maxn], t, cnt[maxn];
struct query {
  long long id, l, r;
} q[maxn];
bool cmp(query a, query b) {
  if ((a.l / sq) == (b.l / sq)) return a.r < b.r;
  return a.l < b.l;
}
void add(long long x) {
  if (cnt[ab[x]] == x) res--;
  cnt[ab[x]]++;
  if (cnt[ab[x]] == x) res++;
}
void del(long long x) {
  if (cnt[ab[x]] == x) res--;
  cnt[ab[x]]--;
  if (cnt[ab[x]] == x) res++;
}
int main() {
  cin >> n >> m;
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
    if (ab[arr[i]] == 0) {
      t++;
      ab[arr[i]] = t;
      ba[t] = arr[i];
    }
  }
  for (int i = 0; i < m; i++) {
    cin >> q[i].l >> q[i].r;
    q[i].id = i;
    q[i].l--;
  }
  sort(q, q + m, cmp);
  long long L = 0, R = 0;
  for (int i = 0; i < m; i++) {
    while (R < q[i].r) add(arr[R++]);
    while (L > q[i].l) add(arr[--L]);
    while (R > q[i].r) del(arr[--R]);
    while (L < q[i].l) del(arr[L++]);
    ans[q[i].id] = res;
  }
  for (int i = 0; i < m; i++) cout << ans[i] << endl;
}
