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;
long block_size;
bool comp(vector<long> a, vector<long> b) {
  if ((a[0] / block_size) == (b[0] / block_size)) return (a[1] < b[1]);
  return (a[0] < b[0]);
}
int main() {
  cin.tie(0);
  cout.tie(0);
  ios_base::sync_with_stdio(0);
  long n, m, s = 0;
  cin >> n >> m;
  block_size = sqrt(n);
  vector<long long> a(n + 1, 0);
  vector<vector<long> > b(m, vector<long>(3));
  vector<long> k(m, 0), sl(1e5 + 1, 0), d(1e5 + 1, 0);
  for (long i = 1; i <= n; cin >> a[i], ++i)
    ;
  for (long i = 0; i < m; cin >> b[i][0] >> b[i][1], b[i][2] = i, ++i)
    ;
  sort(b.begin(), b.end(), comp);
  for (long j = b[0][0]; j <= b[0][1]; ++j)
    if (a[j] <= 1e5) {
      if ((++sl[a[j]]) == a[j])
        ++s, d[a[j]] = 1;
      else if (sl[a[j]] > a[j] && d[a[j]])
        --s, d[a[j]] = 0;
    }
  k[b[0][2]] = s;
  for (long i = 1; i < b.size(); ++i) {
    if (b[i - 1][0] < b[i][0]) {
      for (long j = b[i - 1][0]; j < b[i][0]; ++j)
        if (a[j] <= 1e5) {
          if ((--sl[a[j]]) == a[j])
            ++s, d[a[j]] = 1;
          else if (sl[a[j]] < a[j] && d[a[j]])
            --s, d[a[j]] = 0;
        }
    } else {
      for (long j = b[i][0]; j < b[i - 1][0]; ++j)
        if (a[j] <= 1e5) {
          if ((++sl[a[j]]) == a[j])
            ++s, d[a[j]] = 1;
          else if (sl[a[j]] > a[j] && d[a[j]])
            --s, d[a[j]] = 0;
        }
    }
    if (b[i - 1][1] < b[i][1]) {
      for (long j = b[i - 1][1] + 1; j <= b[i][1]; ++j)
        if (a[j] <= 1e5) {
          if ((++sl[a[j]]) == a[j])
            ++s, d[a[j]] = 1;
          else if (sl[a[j]] > a[j] && d[a[j]])
            --s, d[a[j]] = 0;
        }
    } else {
      for (long j = b[i][1] + 1; j <= b[i - 1][1]; ++j)
        if (a[j] <= 1e5) {
          if ((--sl[a[j]]) == a[j])
            ++s, d[a[j]] = 1;
          else if (sl[a[j]] < a[j] && d[a[j]])
            --s, d[a[j]] = 0;
        }
    }
    k[b[i][2]] = s;
  }
  for (long i = 0; i < k.size(); cout << k[i] << '\n', ++i)
    ;
}
