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::collections::*;
use std::io::*;
use std::*;
// template {{{
macro_rules! get {
    ($($t:tt),*; $n:expr) => {
        {
            let stdin = std::io::stdin();
            let ret = std::io::BufRead::lines(stdin.lock()).take($n).map(|line| {
                let line = line.unwrap();
                let mut it = line.split_whitespace();
                _get!(it; $($t),*)
            }).collect::<Vec<_>>();
            ret
        }
    };
    ($($t:tt),*) => {
        {
            let mut line = String::new();
            std::io::stdin().read_line(&mut line).unwrap();
            let mut it = line.split_whitespace();
            _get!(it; $($t),*)
        }
    };
}
macro_rules! _get {
    ($it:ident; [char]) => {
        _get!($it; String).chars().collect::<Vec<_>>()
    };
    ($it:ident; [u8]) => {
        _get!($it; String).bytes().collect::<Vec<_>>()
    };
    ($it:ident; usize1) => {
        $it.next().unwrap().parse::<usize>().unwrap_or_else(|e| panic!("{}", e)) - 1
    };
    ($it:ident; [usize1]) => {
        $it.map(|s| s.parse::<usize>().unwrap_or_else(|e| panic!("{}", e)) - 1).collect::<Vec<_>>()
    };
    ($it:ident; [$t:ty]) => {
        $it.map(|s| s.parse::<$t>().unwrap_or_else(|e| panic!("{}", e))).collect::<Vec<_>>()
    };
    ($it:ident; $t:ty) => {
        $it.next().unwrap().parse::<$t>().unwrap_or_else(|e| panic!("{}", e))
    };
    ($it:ident; $($t:tt),+) => {
        ($(_get!($it; $t)),*)
    };
}
// }}}

fn main() {
    let n = get!(usize);
    let l = get!([i64]);
    let mut ans = 0;
    for i in 0..n {
        for j in 0..n {
            if i >= j {
                continue;
            }
            if l[i] == l[j] {
                continue;
            }
            for k in 0..n {
                if j >= k {
                    continue;
                }
                if l[j] == l[k] || l[i] == l[k] {
                    continue;
                }
                let m = l[i].max(l[j].max(l[k]));
                let s = l[i] + l[j] + l[k];
                if s-m > m {
                    ans += 1;
                }
            }
        }
    }
    println!("{}", ans);
}