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 JavaScript (Node.js 12.16.1) (judge verdict: Wrong Answer)

function Main(input) {
	const col = input.split("\n");
	const row = col[1].split(" ");

	if(row.length < 3){
		console.log(0);
	} else {
		const triangle = function(a, b, c) {
			const check = (d, e ,f) => (d + e) > f;
			return check(a, b, c) && check(b, c, a) && check(c, a, b);
		}
		const diff = (a, b, c) => a !== b && b !== c && c !== a;

		let list = [];
		for(let s=0; s < row.length; s++){
			list.push(parseInt(row[s], 10));
		}
		let cnt = 0;
		for(let i=0; i < list.length-2; i++){
			for(let j=i+1; j < list.length-1; j++){
				for(let k=j+1; k < list.length; k++){
					if(triangle(list[i],list[j],list[k]) && diff(list[i],list[j],list[k])) {
					console.log(`(${i+1},${j+1},${k+1})`);
						cnt++;
					}
				}
			}
		}
		
		console.log(cnt);
	}
};

Main(require("fs").readFileSync("/dev/stdin", "utf8"));