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()
{
      int n;
      scanf("%d",&n);
      int a[n],i;
      for(i=0;i<n;i++){
            scanf("%d",&a[i]);
      }
      int j,k,x=0,ans=0,tm;
      for(i=0;i<n;i++){
            for(j=i+1;j<n;j++){
                  if(a[i]>a[j]){
                        tm=a[i];
                        a[i]=a[j];
                        a[j]=tm;
                  }
            }
      }
      for(i=0;i<n-2;i++){
            if(i>0 && a[i]==a[i-1]){
                  ans+=x;
                  ///printf("x = %d ans= %d \n",x,ans);
                  continue;
            }
            x=0;
            for(j=i+1;j<n;j++){
                  if(a[i]==a[j]) continue;
                  for(k=j+1;k<n;k++){
                        if(a[k]==a[j]) continue;
                        if(a[i]+a[j]>a[k] && a[i]+a[k]>a[j] && a[j]+a[k]>a[i]) x++;
                  }
            }
            ans+=x;
           /// printf("x = %d ans= %d \n",x,ans);
      }
      printf("%d\n",ans);
}