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: Wrong Answer)

import java.util.Scanner;

public class Main {
	 public static void main(String args[]){
	    	Scanner scanner = new Scanner(System.in);

	    	int n=scanner.nextInt();
	    	Integer[] m=new Integer[n];
	    	int t=0;

	    	int a=0;
	    	int b=0;
	    	int c=0;


	    	for(int h=0;h<n;h++) {
	    		m[h]=scanner.nextInt();
	    	}

	    	for(int j=0;j<n;j++) {
	    		a=m[j];
	    		for(int i=0;i<n;i++) {
	    			b=m[i];
	    			for(int k=0;k<n;k++) {
	    				c=m[k];
	    		if(a!=b && a!=c) {

	    				if(c<b) {
	    					int s=c;
	    					c=b;
	    					b=s;
	    			}
	    				if(c<a) {
	    					int s=c;
	    					c=a;
	    					a=s;
	    				}
	    				if(b<a) {
	    					int s=b;
	    					b=a;
	    					a=s;
	    				}
	    			if((a+b)<c && (c-a-b)<0) {
	    				t=t+1;
	    			}

	    		}
	    	}}}





	    	 }
}