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;
template <class T>
istream& operator>>(istream& is, vector<T>& input) {
  for (T& in : input) is >> in;
  return is;
}
namespace Mo {
int k = 345, l = 0, r = -1, n, m;
long long sum = 0;
vector<int> input;
int cnt[(int)1e5 + 1] = {0};
vector<long long> ans;
struct info {
  int l, r, id;
  info() : l(0), r(0), id(0) {}
  info(int _l, int _r, int _id) : l(_l), r(_r), id(_id) {}
  bool operator<(const info& a) const {
    if (a.l / k == l / k) return a.r > r;
    return a.l > l;
  }
};
vector<info> query;
inline void add(int ind) {
  if (input[ind] > n) return;
  cnt[input[ind]]++;
  if (cnt[input[ind]] == input[ind]) sum++;
  if (cnt[input[ind]] == input[ind] + 1) sum--;
}
inline void remove(int ind) {
  if (input[ind] > n) return;
  cnt[input[ind]]--;
  if (cnt[input[ind]] == input[ind]) sum++;
  if (cnt[input[ind]] == input[ind] - 1) sum--;
}
}  // namespace Mo
using namespace Mo;
void solve() {
  cin >> n >> m;
  input.resize(n);
  cin >> input;
  k = 4 * sqrt(n);
  for (int i = 0; i < m; i++) {
    int u, v;
    cin >> u >> v;
    query.emplace_back(--u, --v, i);
    ans.push_back(0);
  }
  sort(query.begin(), query.end());
  for (int i = 0; i < m; i++) {
    while (r < query[i].r) add(++r);
    while (r > query[i].r) remove(r--);
    while (l < query[i].l) remove(l++);
    while (l > query[i].l) add(--l);
    ans[query[i].id] = sum;
  }
  for (auto it : ans) cout << it << "\n";
}
int main() {
  int tc = 1;
  while (tc--) {
    solve();
  }
  return 0;
}
