#include <bits/stdc++.h>
using namespace std;
long long int n;
long long int a[123456];
map<long long int, long long int> ct;
int main() {
  cin >> n;
  for (long long int i = 0; i < n; i++) cin >> a[i];
  sort(a, a + n);
  long long int ans = a[0] * a[1] * a[2];
  for (long long int i = 0; i < n; i++) ct[a[i]]++;
  int fans = 0;
  if (a[0] == a[1] && a[1] == a[2]) {
    cout << ct[a[0]] * (ct[a[0]] - 1) * (ct[a[0]] - 2) / 6 << endl;
  } else if (a[1] == a[2]) {
    cout << ct[a[1]] * (ct[a[1]] - 1) / 2 << endl;
  } else {
    cout << ct[a[2]] << endl;
  }
  return 0;
}

### Prompt

Create a solution in cpp for 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;
long long int n;
long long int a[123456];
map<long long int, long long int> ct;
int main() {
  cin >> n;
  for (long long int i = 0; i < n; i++) cin >> a[i];
  sort(a, a + n);
  long long int ans = a[0] * a[1] * a[2];
  for (long long int i = 0; i < n; i++) ct[a[i]]++;
  int fans = 0;
  if (a[0] == a[1] && a[1] == a[2]) {
    cout << ct[a[0]] * (ct[a[0]] - 1) * (ct[a[0]] - 2) / 6 << endl;
  } else if (a[1] == a[2]) {
    cout << ct[a[1]] * (ct[a[1]] - 1) / 2 << endl;
  } else {
    cout << ct[a[2]] << endl;
  }
  return 0;
}

```