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 data {
  int l, r, id;
};
int n, m, S, ans;
vector<data> query;
int ans_query[1000000], a[1000000], cnt[1000000];
bool operator<(data x, data y) {
  if (x.r / S != y.r / S) return (x.r / S) < (y.r / S);
  return x.l < y.l;
}
void add(int x) {
  if (x > n || x == 0) return;
  if (cnt[x] == x) ans--;
  cnt[x]++;
  if (cnt[x] == x) ans++;
}
void sub(int x) {
  if (x > n || x == 0) return;
  if (cnt[x] == x) ans--;
  cnt[x]--;
  if (cnt[x] == x) ans++;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m;
  S = sqrt(n);
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  for (int i = 1; i <= m; i++) {
    int l, r;
    cin >> l >> r;
    query.push_back({l, r, i});
  }
  sort(query.begin(), query.end());
  int L = 0, R = 0;
  for (int i = 0; i < query.size(); i++) {
    int l = query[i].l, r = query[i].r, pos = query[i].id;
    while (R < r) add(a[++R]);
    while (L < l) sub(a[L++]);
    while (R > r) sub(a[R--]);
    while (L > l) add(a[--L]);
    ans_query[pos] = ans;
  }
  for (int i = 1; i <= m; i++) cout << ans_query[i] << "\n";
  return 0;
}
