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 proconio::{fastout, input};
use itertools::Itertools;

#[fastout]
fn main() {
    input! {
        n: usize,
        l: [u64; n],
    }

    if n < 3 {
        println!("{}", 0);
        return;
    }

    let mut count = 0;

    for i in 0..n-2 {
    for j in i+1..n-1 {
    for k in j+1..n {
        if l[i] == l[j] { continue; }
        if l[j] == l[k] { continue; }
        if l[k] == l[i] { continue; }
        let comb: Vec<u64> = vec![l[i], l[j], l[k]];
        if comb.iter().sum::<u64>() <= (*comb.iter().max().unwrap()) * 2 { continue; }
        count += 1;
    }

    }
        /* l[i..i+3].iter().permutations(3).for_each(|comb| {
            let mut st_comb = comb.copy();
            if *comb[0] + *comb[1] > *comb[2] {
                count += 1;
            }
        }) */
    }
    /* for _ in 1..=l.len() {
        for i in 0..l.len()-3 {
            if l[i] + l[i+1] > l[i+2] {
                count += 1;
            } else if l[i+1] + l[i+2] > l[i] {
                count += 1;
            } else if l[i+2] + l[i] > l[i+1] {
                count += 1;
            }
        }
 */
        /* let tmp = l[0];
        l.drain(..1);
        l.push(tmp); */
    

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