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 (Clang 10.0.0) (judge verdict: Accepted)

#include <stdio.h>

int sankaku (int o,int p,int s)
{
	int max;
	if(o==p||p==s||s==o){
		return 0;
	}
	if(o>p&&o>s){
		max=o;
		if(max>=(p+s)){
			return 0;
		}
	}else if(p>o&&p>s){
		max=p;
		if(max>=(o+s)){
			return 0;
		}
	}else{
		max=s;
		if(max>=(p+o)){
			return 0;
		}
	}
	return 1;
}

	
 
int main (void)
{
	int n,i,k,q,count=0;
	int l[100];
	scanf("%d",&n);
	
	for(i=0;i<n;i++){
		scanf("%d",&l[i]);
	}
	
	if(n<=2){
		printf("%d",count);
		return 0;
	}
	for(i=0;i<n;i++){
		for(k=0;k<n;k++){
			for(q=0;q<n;q++){
				if(i!=k&&k!=q&&q!=i){
					if(sankaku(l[i],l[k],l[q])==1){
						count++;
					}
					
				}
			}
		}
	}
	printf("%d\n",count/6);
	return 0;
}