#include <bits/stdc++.h>
using namespace std;
long long int ncr(long long int n, long long int r) {
  if (r == 1)
    return n;
  else
    return (ncr(n - 1, r - 1) * n / r);
}
int main() {
  long long int i, n, c1 = 0, c2, c3;
  cin >> n;
  int ar[n];
  for (auto i = (0); i < (n); i++) cin >> ar[i];
  sort(ar, ar + n);
  while (ar[c1] == ar[0] && c1 < n) c1++;
  c2 = c1;
  while (ar[c2] == ar[c1] && c2 < n) c2++;
  c3 = c2;
  while (ar[c3] == ar[c2] && c3 < n) c3++;
  if (c1 >= 3)
    cout << ncr(c1, 3) << endl;
  else if (c1 == 2)
    cout << c2 - c1 << endl;
  else if (c2 - c1 >= 2)
    cout << ncr(c2 - c1, 2) << endl;
  else
    cout << c3 - c2 << endl;
  return 0;
}

### Prompt

Your challenge is to write 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;
long long int ncr(long long int n, long long int r) {
  if (r == 1)
    return n;
  else
    return (ncr(n - 1, r - 1) * n / r);
}
int main() {
  long long int i, n, c1 = 0, c2, c3;
  cin >> n;
  int ar[n];
  for (auto i = (0); i < (n); i++) cin >> ar[i];
  sort(ar, ar + n);
  while (ar[c1] == ar[0] && c1 < n) c1++;
  c2 = c1;
  while (ar[c2] == ar[c1] && c2 < n) c2++;
  c3 = c2;
  while (ar[c3] == ar[c2] && c3 < n) c3++;
  if (c1 >= 3)
    cout << ncr(c1, 3) << endl;
  else if (c1 == 2)
    cout << c2 - c1 << endl;
  else if (c2 - c1 >= 2)
    cout << ncr(c2 - c1, 2) << endl;
  else
    cout << c3 - c2 << endl;
  return 0;
}

```