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;
#pragma GCC optimize("O3")
int n, q, blk = 500;
long long ans = 0;
long long arr[200005];
struct query {
  int l, r, i;
};
query Q[200005];
long long A[200005];
unordered_map<long long, long long> ct;
bool cmp(query A, query B) {
  if (A.l / blk != B.l / blk) return A.l / blk < B.l / blk;
  return A.r < B.r;
}
inline void add(int pos) {
  ct[arr[pos]]++;
  if (ct[arr[pos]] == arr[pos])
    ans++;
  else if (ct[arr[pos]] == arr[pos] + 1)
    ans--;
}
inline void remove(int pos) {
  ct[arr[pos]]--;
  if (ct[arr[pos]] == arr[pos])
    ans++;
  else if (ct[arr[pos]] == arr[pos] - 1)
    ans--;
}
void solve() {
  cin >> n >> q;
  for (long long i = 0; i < n; i++) cin >> arr[i];
  for (long long i = 0; i < q; i++) {
    cin >> Q[i].l >> Q[i].r;
    Q[i].l--;
    Q[i].r--;
    Q[i].i = i;
  }
  sort(Q, Q + q, cmp);
  long long l = 0, r = 0;
  for (long long i = 0; i < q; i++) {
    while (l > Q[i].l) {
      add(l - 1);
      l--;
    }
    while (r <= Q[i].r) {
      add(r);
      r++;
    }
    while (l < Q[i].l) {
      remove(l);
      l++;
    }
    while (r > Q[i].r + 1) {
      remove(r - 1);
      r--;
    }
    A[Q[i].i] = ans;
  }
  for (long long i = 0; i < q; i++) cout << A[i] << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t = 1;
  for (long long a = 0; a < t; a++) {
    solve();
  }
}
