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 main() {
  static const int MAXN = 100000;
  int N, M;
  scanf("%d %d", &N, &M);
  static int a[MAXN];
  map<int, int> m;
  for (int i = 0; i < N; i++) {
    scanf("%d", &a[i]);
    ++m[a[i]];
  }
  static int ls[MAXN], rs[MAXN];
  for (int i = 0; i < M; i++) {
    scanf("%d %d", &ls[i], &rs[i]);
    --ls[i];
    --rs[i];
  }
  vector<int> candidates;
  for (map<int, int>::const_iterator it = m.begin(); it != m.end(); ++it) {
    if (it->first <= it->second) {
      candidates.push_back(it->first);
    }
  }
  static int ans[MAXN];
  for (vector<int>::const_iterator it = candidates.begin();
       it != candidates.end(); ++it) {
    static int sums[MAXN + 1];
    for (int i = 0; i < N; i++) {
      sums[i + 1] = sums[i] + (a[i] == *it);
    }
    for (int i = 0; i < M; i++) {
      const int cnt = sums[rs[i] + 1] - sums[ls[i]];
      if (cnt == *it) {
        ++ans[i];
      }
    }
  }
  for (int i = 0; i < M; i++) {
    printf("%d\n", ans[i]);
  }
  return 0;
}
