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 PHP (7.4.4) (judge verdict: Wrong Answer)

<?php
// Your code here!
    $n = trim(fgets(STDIN));
    
    $input = explode(" ", trim(fgets(STDIN)));
    
    $array = array_unique($input);
    
    rsort($array);
    
    $count = 0;
    
    foreach($array as $key1 => $first){
        if(!isset($array[$key1 + 2]) || $first >= $array[$key1 + 1] + $array[$key1 + 2]){
            break;
        }
        
            
        foreach(array_slice($array, $key1 + 1, null, true) as $key2 => $second){
            if(!isset($array[$key2 + 1]) || $first >= $second + $array[$key2 + 1]){
                break;
            }
            foreach(array_slice($array, $key2 + 1, null,  true) as $third){
                if($first >= $second + $third){
                    break;
                }
                
                
                $options[] = count(array_keys($input, $first));
                $options[] = count(array_keys($input, $second));
                $options[] = count(array_keys($input, $third));
                
                $count += array_product($options);
                
                unset($options);
                
                //echo "($first, $second, $third)$count\n";
                
            }
        }
    }
    
    echo $count;
    
?>