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: Wrong Answer)

#include <stdio.h>

int main () {
    int n;
    long long int L[101];
    scanf("%d",&n);
    for (int i=0; i<n; i++) {
        scanf("%lld",L + i);
    }
    int ans = 0;
    long long int tmp, sum;

    for(int i=0; i<n; i++) {
        for(int j=i+1; j<n; j++) {
            for (int k=j+2; k<n; k++) {
                if (L[i] != L[j] && L[j] != L[k] && L[k] != L[i]) {
                    // max 
                    if(n<3) break;
                    int max = L[i];
                    if (L[j] > max) max = L[j];
                    if (L[k] > max) max = L[k];
                    sum = L[i] + L[j] + L[k];
                    if (sum - tmp*2 > 0) ans ++;
                }
            }
        }
    }

    printf("%d\n", ans);

    return 0;
}