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 double eps = 1.0e-11;
const double pi = acos(-1.0);
const int N = 100009;
int a[N];
int main() {
  int n, m;
  cin >> n >> m;
  map<int, int> freq;
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
    if (freq.count(a[i]) == 0)
      freq[a[i]] = 1;
    else
      ++freq[a[i]];
  }
  vector<int> good;
  for (map<int, int>::iterator it = freq.begin(); it != freq.end(); ++it)
    if (it->first <= it->second) good.push_back(it->first);
  vector<vector<int> > ch(n + 1, vector<int>(good.size()));
  for (long long i = 1; i <= n; i++)
    for (long long j = 0; j < (int)(ch[i]).size(); j++)
      ch[i][j] = ch[i - 1][j] + (a[i - 1] == good[j]);
  for (long long i = 0; i < m; i++) {
    int l, r;
    cin >> l >> r;
    --l, --r;
    int res = 0;
    for (long long j = 0; j < ch[0].size(); j++) {
      if (ch[r + 1][j] - ch[l][j] == good[j]) ++res;
    }
    cout << res << endl;
  }
  return 0;
}
