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 Java (OpenJDK 11.0.6) (judge verdict: Accepted)

import java.util.*;

public class Main{
  public static void main(String[] args){
    Scanner scan = new Scanner(System.in);
    int count=0,l1,l2,l3;
    int max = scan.nextInt(),tmp;
    int[] bars = new int[max];
    for(int i=0;i<max;i++)bars[i] = scan.nextInt();
    for(int i=0;i<max-2;i++){
      for(int j=0;j<max-i-1;j++){
        if(bars[j]<bars[j+1]){
          tmp=bars[j];
          bars[j]=bars[j+1];
          bars[j+1]=tmp;
        }
      }
    }
    for(int i=0; i<max-2;i++){
      l1=bars[i];
      for(int j=i+1; j<max-1;j++){
        l2=bars[j];
        for(int k=j+1; k<max;k++){
          if(l1==l2)break;
          l3=bars[k];
          if(l1<(l2+l3) && l2<(l1+l3) && l3<(l1+l2)&& l3 != l1 && l3 != l2){
            count++;
          }
        }
      }
    }
    
    System.out.println(""+count);
  }
  
}