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 procon_read<T: std::str::FromStr>() ->  Vec<T> {
    let mut buf = String::new();
    io::stdin().read_line(&mut buf).expect("");
    let iter = buf.split_whitespace();
    let mut v: Vec<T> = Vec::new();
    for i in iter {
        v.push(i.parse().ok().unwrap());
    }
    v
}

fn main() {
    let n:usize = procon_read()[0];
    let mut v:Vec<u64> = procon_read();
    v.sort();
    let mut cnt:u64 = 0;
    if n < 3 {
        
    }else{
        for i in 0..(n-2) {
            for j in (i+1)..(n-1) {
                for k in (j+1)..n {
                    if v[i] != v[j] && v[j] != v[k] && v[i] + v[j] > v[k] {
                        cnt += 1;
                    }
                }
            }
        }
    }
    println!("{}", cnt);
}