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 Rust (1.42.0) (judge verdict: Accepted)

use std::io;

fn input() -> String {
    let mut inp = String::new();
    io::stdin().read_line(&mut inp).unwrap();
    inp.trim().to_string()
}

fn main() {
    let n: usize = input().parse().unwrap();
    let l: Vec<i32> = input()
        .split_whitespace()
        .map(|x| x.parse().unwrap())
        .collect();

    let mut ans = 0;
    for i in 0..n {
        for j in (i + 1)..n {
            for k in (j + 1)..n {
                if l[i] == l[j] || l[j] == l[k] || l[i] == l[k] {
                    continue;
                }
                if l[i] + l[j] > l[k] && l[j] + l[k] > l[i] && l[i] + l[k] > l[j] {
                    ans += 1;
                }
            }
        }
    }

    println!("{}", ans);
}