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 int N = 100005;
int mp[N];
int ans;
int SEGMENT;
bool mo(pair<pair<int, int>, int> a, pair<pair<int, int>, int> b) {
  if (a.first.first / SEGMENT == b.first.first / SEGMENT)
    return a.first.second < b.first.second;
  return a.first.first / SEGMENT < b.first.first / SEGMENT;
}
void add(int x) {
  if (x >= N) return;
  if (mp[x] == x) ans--;
  mp[x]++;
  if (mp[x] == x) ans++;
}
void remove(int x) {
  if (x >= N) return;
  if (mp[x] == x) ans--;
  mp[x]--;
  if (mp[x] == x) ans++;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n, t;
  cin >> n >> t;
  int a[n + 1];
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  SEGMENT = sqrt(n);
  pair<pair<int, int>, int> p[t];
  for (int i = 0; i < t; i++) {
    cin >> p[i].first.first >> p[i].first.second;
    p[i].second = i;
  }
  sort(p, p + t, mo);
  int q[t];
  int l = 1;
  int r = 0;
  for (int i = 0; i < t; i++) {
    while (l < p[i].first.first) {
      remove(a[l]);
      l++;
    }
    while (l > p[i].first.first) {
      l--;
      add(a[l]);
    }
    while (r < p[i].first.second) {
      r++;
      add(a[r]);
    }
    while (r > p[i].first.second) {
      remove(a[r]);
      r--;
    }
    q[p[i].second] = ans;
  }
  for (int i = 0; i < t; i++) {
    cout << q[i] << "\n";
  }
}
