#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  long long int n;
  cin >> n;
  long long int a[n];
  long long int b, c, d, i, j, k;
  for (i = 0; i < n; i++) {
    cin >> a[i];
  }
  sort(a, a + n);
  b = 0;
  i = 3;
  while (a[i] == a[2] && i < n) b++, i++;
  if (a[0] == a[2]) {
    b += 3;
    b = b * (b - 1) * (b - 2);
    b /= 6;
  } else if (a[1] == a[2]) {
    b += 2;
    b = b * (b - 1);
    b /= 2;
  } else
    b++;
  cout << b;
  return 0;
}

### Prompt

Generate a CPP solution 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;
int main() {
  ios_base::sync_with_stdio(false);
  long long int n;
  cin >> n;
  long long int a[n];
  long long int b, c, d, i, j, k;
  for (i = 0; i < n; i++) {
    cin >> a[i];
  }
  sort(a, a + n);
  b = 0;
  i = 3;
  while (a[i] == a[2] && i < n) b++, i++;
  if (a[0] == a[2]) {
    b += 3;
    b = b * (b - 1) * (b - 2);
    b /= 6;
  } else if (a[1] == a[2]) {
    b += 2;
    b = b * (b - 1);
    b /= 2;
  } else
    b++;
  cout << b;
  return 0;
}

```