#include <bits/stdc++.h>
using namespace std;
void solve() {
  long long int ans = 0;
  long long int n;
  cin >> n;
  vector<long long int> a(n);
  for (long long int i = 0; i < n; i++) {
    cin >> a[i];
  }
  sort(a.begin(), a.end());
  for (long long int i = 0; i < n; i++)
    if (a[i] == a[2]) ans++;
  if (a[2] == a[0])
    ans = ans * (ans - 1) * (ans - 2) / 6;
  else if (a[2] == a[1])
    ans = ans * (ans - 1) / 2;
  cout << ans << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int t;
  t = 1;
  while (t--) {
    solve();
  }
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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;
void solve() {
  long long int ans = 0;
  long long int n;
  cin >> n;
  vector<long long int> a(n);
  for (long long int i = 0; i < n; i++) {
    cin >> a[i];
  }
  sort(a.begin(), a.end());
  for (long long int i = 0; i < n; i++)
    if (a[i] == a[2]) ans++;
  if (a[2] == a[0])
    ans = ans * (ans - 1) * (ans - 2) / 6;
  else if (a[2] == a[1])
    ans = ans * (ans - 1) / 2;
  cout << ans << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int t;
  t = 1;
  while (t--) {
    solve();
  }
}

```