Problem:
Score : 200 points

Problem Statement

We have sticks numbered 1, \cdots, N. The length of Stick i (1 \leq i \leq N) is L_i.

In how many ways can we choose three of the sticks with different lengths that can form a triangle?

That is, find the number of triples of integers (i, j, k) (1 \leq i < j < k \leq N) that satisfy both of the following conditions:

L_i, L_j, and L_k are all different.

There exists a triangle whose sides have lengths L_i, L_j, and L_k.

Constraints

1 \leq N \leq 100

1 \leq L_i \leq 10^9

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
L_1 L_2 \cdots L_N

Output

Print the number of ways to choose three of the sticks with different lengths that can form a triangle.

Sample Input 1

5
4 4 9 7 5

Sample Output 1

5

The following five triples (i, j, k) satisfy the conditions: (1, 3, 4), (1, 4, 5), (2, 3, 4), (2, 4, 5), and (3, 4, 5).

Sample Input 2

6
4 5 4 3 3 5

Sample Output 2

8

We have two sticks for each of the lengths 3, 4, and 5. To satisfy the first condition, we have to choose one from each length.

There is a triangle whose sides have lengths 3, 4, and 5, so we have 2 ^ 3 = 8 triples (i, j, k) that satisfy the conditions.

Sample Input 3

10
9 4 6 1 9 6 10 6 6 8

Sample Output 3

39

Sample Input 4

2
1 1

Sample Output 4

0

No triple (i, j, k) satisfies 1 \leq i < j < k \leq N, so we should print 0.

Solution in C# (.NET Core 3.1.201) (judge verdict: Runtime Error)

using System;
class Program
{
    static void Main(string[] args)
    {
        int i, j=0;
        int N = int.Parse(Console.ReadLine());
        string[] input = Console.ReadLine().Split(' ');
        int[] L=new int[N];
        int[] l=new int[N];
        for(i=0; i<N; i++){
            L[i]=int.Parse(input[i]);
            if(l[L[i]]==0){
                l[L[i]]=1;
                j++;
            }
        }
        if(j<3)
            Console.WriteLine(0);
        else
            Console.WriteLine(Factorial(j)/(6*Factorial(j-3)));
    }

    static long Factorial(int n) =>
    n == 0 ? 1L : n * Factorial(n - 1);
}