#include <bits/stdc++.h>
using namespace std;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  long long int n;
  cin >> n;
  map<long long int, long long int> g;
  for (long long int i = 0; i < n; i++) {
    long long int z;
    cin >> z;
    g[z]++;
  }
  long long int fir_siz = g.begin()->second;
  if (fir_siz >= 3) {
    cout << fir_siz * (fir_siz - 1) * (fir_siz - 2) / 6ll;
  } else if (fir_siz == 2) {
    if (g.size() == 1) {
      cout << 0;
      return 0;
    }
    cout << (++g.begin())->second;
  } else {
    if (g.size() == 1) {
      cout << 0;
      return 0;
    }
    if (g.size() == 2) {
      long long int sec_siz = (++g.begin())->second;
      cout << sec_siz * (sec_siz - 1) / 2ll;
    } else {
      long long int sec_siz = (++g.begin())->second;
      if (sec_siz == 1) {
        map<long long int, long long int>::iterator it = g.begin();
        ++it;
        ++it;
        long long int third_siz = it->second;
        cout << third_siz;
      } else {
        cout << sec_siz * (sec_siz - 1) / 2ll;
      }
    }
  }
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

After returning from the army Makes received a gift — an array a consisting of n positive integer numbers. He hadn't been solving problems for a long time, so he became interested to answer a particular question: how many triples of indices (i, j, k) (i < j < k), such that ai·aj·ak is minimum possible, are there in the array? Help him with it!

Input

The first line of input contains a positive integer number n (3 ≤ n ≤ 105) — the number of elements in array a. The second line contains n positive integer numbers ai (1 ≤ ai ≤ 109) — the elements of a given array.

Output

Print one number — the quantity of triples (i, j, k) such that i, j and k are pairwise distinct and ai·aj·ak is minimum possible.

Examples

Input

4
1 1 1 1


Output

4


Input

5
1 3 2 3 4


Output

2


Input

6
1 3 3 1 3 2


Output

1

Note

In the first example Makes always chooses three ones out of four, and the number of ways to choose them is 4.

In the second example a triple of numbers (1, 2, 3) is chosen (numbers, not indices). Since there are two ways to choose an element 3, then the answer is 2.

In the third example a triple of numbers (1, 1, 2) is chosen, and there's only one way to choose indices.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  long long int n;
  cin >> n;
  map<long long int, long long int> g;
  for (long long int i = 0; i < n; i++) {
    long long int z;
    cin >> z;
    g[z]++;
  }
  long long int fir_siz = g.begin()->second;
  if (fir_siz >= 3) {
    cout << fir_siz * (fir_siz - 1) * (fir_siz - 2) / 6ll;
  } else if (fir_siz == 2) {
    if (g.size() == 1) {
      cout << 0;
      return 0;
    }
    cout << (++g.begin())->second;
  } else {
    if (g.size() == 1) {
      cout << 0;
      return 0;
    }
    if (g.size() == 2) {
      long long int sec_siz = (++g.begin())->second;
      cout << sec_siz * (sec_siz - 1) / 2ll;
    } else {
      long long int sec_siz = (++g.begin())->second;
      if (sec_siz == 1) {
        map<long long int, long long int>::iterator it = g.begin();
        ++it;
        ++it;
        long long int third_siz = it->second;
        cout << third_siz;
      } else {
        cout << sec_siz * (sec_siz - 1) / 2ll;
      }
    }
  }
  return 0;
}

```