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>
#pragma optimize("Ofast")
using namespace std;
using ll = long long;
const int N = 1e5 + 5;
int n, l, r, query;
int a[N], BLOCK;
vector<pair<pair<int, int>, int>> q(N);
bool cmp(pair<pair<int, int>, int> a, pair<pair<int, int>, int> b) {
  if (a.first.first / BLOCK != b.first.first / BLOCK) {
    return a.first.first / BLOCK < b.first.first / BLOCK;
  }
  return a.first.second < b.first.second;
}
unordered_map<ll, int> cnt;
int ans;
void add(int pos) {
  if (cnt[a[pos]] == a[pos]) ans--;
  cnt[a[pos]]++;
  if (cnt[a[pos]] == a[pos]) ans++;
}
void delet(int pos) {
  if (cnt[a[pos]] == a[pos]) ans--;
  cnt[a[pos]]--;
  if (cnt[a[pos]] == a[pos]) ans++;
}
void solve() {
  vector<int> res(query);
  int cur_l = 0, cur_r = 0;
  for (int i = 0; i < query; ++i) {
    l = q[i].first.first;
    r = q[i].first.second;
    while (l < cur_l) {
      add(cur_l - 1);
      cur_l--;
    }
    while (r >= cur_r) {
      add(cur_r);
      cur_r++;
    }
    while (l > cur_l) {
      delet(cur_l);
      cur_l++;
    }
    while (r + 1 < cur_r) {
      delet(cur_r - 1);
      cur_r--;
    }
    res[q[i].second] = ans;
  }
  for (int i = 0; i < query; i++) {
    printf("%d\n", res[i]);
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n >> query;
  BLOCK = sqrt(n + .0) + 1;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    if (a[i] > n) a[i] = -1;
  }
  for (int i = 0; i < query; i++) {
    cin >> l >> r;
    q[i] = {{--l, --r}, i};
  }
  sort(q.begin(), q.begin() + query, cmp);
  solve();
  return 0;
}
