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;
const int MAX_LEN = (int)1e5;
int xs[MAX_LEN];
int nums[500][MAX_LEN + 1];
int main() {
  int n, m;
  cin >> n >> m;
  map<int, int> cnt;
  for (int i = 0; i < n; i++) {
    cin >> xs[i];
    cnt[xs[i]]++;
  }
  vector<int> vs;
  for (map<int, int>::iterator itr = cnt.begin(); itr != cnt.end(); itr++) {
    if (itr->second >= itr->first) vs.push_back(itr->first);
  }
  int l = vs.size();
  memset(nums, 0, sizeof(nums));
  for (int i = 0; i < l; i++) {
    for (int j = 0; j < n; j++) {
      nums[i][j + 1] = nums[i][j];
      if (vs[i] == xs[j]) nums[i][j + 1]++;
    }
  }
  for (int i = 0; i < m; i++) {
    int a, b;
    cin >> a >> b;
    int ret = 0;
    for (int j = 0; j < l; j++) {
      if (nums[j][b] - nums[j][a - 1] == vs[j]) ++ret;
    }
    cout << ret << endl;
  }
  return 0;
}
