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 main(void)
{
    int input=0;
    long int number[105]={};
    long int a=0;
    long int b=0;
    long int c=0;
    int count=0;
    
    scanf("%d",&input);
    for(int i = 0; i < input; i++){
        scanf("%ld",&number[i]);
    }
    if(input < 3){
        printf("0");
        return 0;
    }
 
    for(int i = 0; i <input; i++){
        for(int j = i + 1; j < input ;j++){
            if(number[i] != number[j]){
                for(int k = j + 1; k < input; k++){
                    if(number[i] != number[k] && number[j] != number[k]){
                        a = number[i]+number[j]-number[k];
                        b = number[j]+number[k]-number[i];
                        c = number[k]+number[i]-number[j];
                        if(a > 0 && b> 0 && c > 0){
                            count+=1;

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