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 block;
unordered_map<int, int> x;
vector<long long> a;
struct query {
  int l, r, pos;
  bool operator<(const query &other) const {
    if (l / block == other.l / block) return r < other.r;
    return l / block < other.l / block;
  }
};
int l = 0, r = -1, cnt = 0;
void remove(int val) {
  if (x[val] == val) cnt--;
  x[val]--;
  if (x[val] == val) cnt++;
  if (x[val] == 0) x.erase(val);
}
void add(int val) {
  if (x[val] == val) cnt--;
  x[val]++;
  if (x[val] == val) cnt++;
}
void solve() {
  long long n, m;
  cin >> n >> m;
  block = sqrt(n) + 1;
  a.resize(n);
  for (long long i = 0; i < n; i++) cin >> a[i];
  vector<query> q(m);
  for (int i = 0; i < m; i++) {
    cin >> q[i].l >> q[i].r;
    q[i].pos = i;
    q[i].l--, q[i].r--;
  }
  sort(q.begin(), q.end());
  vector<int> ans(m);
  for (int i = 0; i < m; i++) {
    while (l < q[i].l) remove(a[l++]);
    while (l > q[i].l) add(a[--l]);
    while (r < q[i].r) add(a[++r]);
    while (r > q[i].r) remove(a[r--]);
    ans[q[i].pos] = cnt;
  }
  for (auto i : ans) cout << i << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  while (t--) solve();
  return 0;
}
