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 ans = 0, n;
int arr[110003];
int couqnt[110003];
int res[110003];
void add(int x) {
  if (arr[x] > n) {
    return;
  }
  if (couqnt[arr[x]] == arr[x]) {
    ans--;
  }
  couqnt[arr[x]]++;
  if (couqnt[arr[x]] == arr[x]) {
    ans++;
  }
}
void remove(int x) {
  if (arr[x] > n) {
    return;
  }
  if (couqnt[arr[x]] == (arr[x])) {
    ans--;
  }
  couqnt[arr[x]]--;
  if (couqnt[arr[x]] == (arr[x])) {
    ans++;
  }
}
int main() {
  int st, en, q;
  int buc = 316;
  cin >> n >> q;
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
  }
  vector<pair<pair<int, int>, pair<int, int> > > v(q);
  for (int i = 0; i < q; i++) {
    cin >> st;
    st--;
    cin >> en;
    v[i] = make_pair(make_pair(st / buc, en), make_pair(st, i));
  }
  sort(v.begin(), v.end());
  int curst = 0;
  int curen = 0;
  for (int i = 0; i < q; i++) {
    st = v[i].second.first;
    en = v[i].first.second;
    int id = v[i].second.second;
    while (curst > st) {
      curst--;
      add(curst);
    }
    while (curen < en) {
      add(curen);
      curen++;
    }
    while (curst < st) {
      remove(curst);
      curst++;
    }
    while (curen > en) {
      curen--;
      remove(curen);
    }
    res[id] = ans;
  }
  for (int i = 0; i < q; i++) {
    printf("%d\n", res[i]);
  }
}
