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 (GCC 9.2.1) (judge verdict: Time Limit Exceeded)

#include<stdio.h>
int main()
{
    int N;
    int L[101] = {};
    int i;
    int index;
    int tmp;

    scanf("%d", &N);
    for ( i = 0; i < N; i++)
    {
        scanf("%d", &L[i]);
        //  sorting
        index = i;
        while (index > 0)
        {
            if(L[index] < L[index - 1])
            {
                tmp = L[index];
                L[index] = L[index - 1];
                L[index - 1] = tmp;
            }
        }
    }
    
    //  counting
    int j, k;
    int counter = 0;

    for (i = 0; i < N; i++)
    {
        for (j = i+1; j < N; j++)
        {
            if(L[i]==L[j])
                continue;

            for ( k = j+1; k < N; k++)
            {
                if (L[j]==L[k])
                    continue;

                if(L[i]+L[j]>L[k])
                {
                    counter += N - k + 1;
                    break;
                }
            }
        }
    }

    printf("%d", counter);
    return 0;
}