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 N = 1e5;
int n, m, arr[N + 1];
vector<int> candiVals, candiValsInd[N + 1];
int main(int argc, char const *argv[]) {
  scanf("%d%d", &n, &m);
  for (int i = 0; i < n; i++) {
    scanf("%d", &arr[i]);
    if (arr[i] <= N) {
      candiValsInd[arr[i]].push_back(i);
    }
  }
  for (int i = 1; i <= N; i++) {
    if (candiValsInd[i].size() >= i) {
      candiVals.push_back(i);
    }
  }
  while (m--) {
    int l, r;
    scanf("%d%d", &l, &r);
    l--;
    r--;
    int ans = 0;
    for (int i = 0; i < candiVals.size(); i++) {
      int val = candiVals[i];
      if (upper_bound(candiValsInd[val].begin(), candiValsInd[val].end(), r) -
              lower_bound(candiValsInd[val].begin(), candiValsInd[val].end(),
                          l) ==
          val) {
        ans++;
      }
    }
    printf("%d\n", ans);
  }
  return 0;
}
