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() {
  int n, m;
  while (scanf("%d%d", &n, &m) >= 1) {
    vector<int> as(n);
    for (int i = 0; i < n; i++) scanf("%d", &as[i]);
    vector<int> vals = as;
    sort(vals.begin(), vals.end());
    vals.erase(unique(vals.begin(), vals.end()), vals.end());
    vector<int> cnt(((int)(vals).size()));
    for (int i = 0; i < n; i++) {
      as[i] = lower_bound(vals.begin(), vals.end(), as[i]) - vals.begin();
      cnt[as[i]]++;
    }
    vector<int> good;
    for (int i = 0; i < ((int)(vals).size()); i++)
      if (cnt[i] >= vals[i]) good.push_back(i);
    vector<vector<int> > cnts(((int)(good).size()), vector<int>(n + 1, 0));
    for (int i = 0; i < ((int)(good).size()); i++) {
      for (int i2 = 0; i2 < n; i2++)
        cnts[i][i2 + 1] = cnts[i][i2] + (as[i2] == good[i]);
    }
    for (int step = 0; step < m; step++) {
      int l, r;
      scanf("%d%d", &l, &r), l--, r--;
      int ans = 0;
      for (int i = 0; i < ((int)(good).size()); i++) {
        int cnt = cnts[i][r + 1] - cnts[i][l];
        ans += cnt == vals[good[i]];
      }
      printf("%d\n", ans);
    }
  }
  return 0;
}
