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;
vector<pair<long long, pair<long long, long long> > > V;
long long BS, answer = 0, a[100000 + 9], cnt[100000 + 9], ans[100000 + 9], n, t;
void add(long long p) {
  if (cnt[a[p]] == a[p]) answer--;
  cnt[a[p]] += 1;
  if (cnt[a[p]] == a[p]) answer++;
}
void remove(long long p) {
  if (cnt[a[p]] == a[p]) answer--;
  cnt[a[p]] -= 1;
  if (cnt[a[p]] == a[p]) answer++;
}
bool cmp(const pair<long long, pair<long long, long long> > &a,
         const pair<long long, pair<long long, long long> > &b) {
  if ((a.second.first) / BS != (b.second.first) / BS)
    return (a.second.first) / BS < (b.second.first) / BS;
  return a.second.second < b.second.second;
}
int main() {
  cin >> n >> t;
  BS = ceil(sqrt(n));
  for (long long i = 0; i < n; i++) cin >> a[i];
  for (long long i = 0; i < t; i++) {
    long long x, y;
    cin >> x >> y;
    V.push_back({i, {x - 1, y - 1}});
  }
  sort(V.begin(), V.end(), cmp);
  long long CL = V[0].second.first, CR = V[0].second.second;
  for (long long i = CL; i <= CR; i++) {
    if (a[i] > 100000) continue;
    add(i);
  }
  ans[V[0].first] = answer;
  for (long long k = 1; k < t; k++) {
    long long L = V[k].second.first, R = V[k].second.second;
    if (CL < L)
      for (long long i = CL; i <= L - 1; i++) {
        if (a[i] > 100000) continue;
        remove(i);
      }
    if (CL > L)
      for (long long i = CL - 1; i >= L; i--) {
        if (a[i] > 100000) continue;
        add(i);
      }
    if (CR > R)
      for (long long i = CR; i >= R + 1; i--) {
        if (a[i] > 100000) continue;
        remove(i);
      }
    if (CR < R)
      for (long long i = CR + 1; i <= R; i++) {
        if (a[i] > 100000) continue;
        add(i);
      }
    ans[V[k].first] = answer;
    CL = L, CR = R;
  }
  for (long long i = 0; i < t; i++) cout << ans[i] << '\n';
  return 0;
}
