#include <bits/stdc++.h>
using namespace std;
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  vector<long long> a;
  map<long long, long long> m;
  long long q, times, tmp;
  cin >> q;
  times = q;
  while (q--) {
    cin >> tmp;
    m[tmp]++;
    a.emplace_back(tmp);
  }
  sort(a.begin(), a.end());
  if (a[1] != a[2]) {
    cout << m[a[2]];
  } else if (a[1] == a[2] && a[0] != a[1]) {
    cout << (m[a[2]] * (m[a[2]] - 1)) / 2;
  } else {
    cout << (m[a[2]] * (m[a[2]] - 1) * (m[a[2]] - 2)) / 6;
  }
}

### Prompt

Please create a solution in Cpp to the following problem:

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;
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  vector<long long> a;
  map<long long, long long> m;
  long long q, times, tmp;
  cin >> q;
  times = q;
  while (q--) {
    cin >> tmp;
    m[tmp]++;
    a.emplace_back(tmp);
  }
  sort(a.begin(), a.end());
  if (a[1] != a[2]) {
    cout << m[a[2]];
  } else if (a[1] == a[2] && a[0] != a[1]) {
    cout << (m[a[2]] * (m[a[2]] - 1)) / 2;
  } else {
    cout << (m[a[2]] * (m[a[2]] - 1) * (m[a[2]] - 2)) / 6;
  }
}

```