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: Accepted)

#include <stdio.h>

int	is_triangle(int a, int b, int c)
{
  int	blen_a;
  int	blen_b;
  int	blen_c;
  
  blen_a = (a < (b + c));
  blen_b = (b < (c + a));
  blen_c = (c < (a + b));
  
  return (blen_a && blen_b && blen_c);
}

int	main(void)
{
  int	n;
  int	*li;
  int	i;
  int	j;
  int	k;
  int	count;
  
  if (scanf("%d", &n) == NULL)
    return (-1);
  if (n < 3)
  {
    printf("0\n");
    return (0);
  }
  if (!(li = (int *)malloc(sizeof(int) * n)))
    return (-1);
  i = 0;
  while (i < n)
  {
    if (scanf("%d", &li[i++]) == NULL)
      return (-1);
  }
  i = 0;
  count = 0;
  while (i < n - 2)
  {
    j = i + 1;
    while (j < n - 1)
    {
      k = j + 1;
      while (k < n)
      {
        if (li[i] != li[j] && li[j] != li[k] && li[k] != li[i])
        {
          if (is_triangle(li[i], li[j], li[k]))
            count++;
        }
        k++;
      }
      j++;
    }
    i++;
  }
  printf("%d\n", count);
  free(li);
  return (0);
}