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)

#![allow(clippy::needless_range_loop)]
#![allow(unused_macros)]
#![allow(dead_code)]
#![allow(unused_imports)]

use itertools::iproduct;
use itertools::Itertools;
use proconio::input;
use proconio::marker::*;

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

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

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