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(unused_imports)]

use std::cmp::{max, min, Ordering};
use std::collections::*;
use std::{i32, i64, i128, u32, u64, u128, f64};
use std::io::{stdout, Write, BufWriter};
use std::mem::{swap};
use proconio::marker::{Isize1, Usize1, Bytes, Chars};

#[proconio::fastout]
fn main() {
    proconio::input!(
        n: usize,
        mut ls: [u32; n],
    );
    let mut counter = HashMap::new();
    for &l in &ls { *counter.entry(l).or_insert(0) += 1; }
    ls.sort();
    ls.dedup();
    if counter.len() < 3 {
        println!("0");
        return;
    }
    let n = ls.len();
    let mut ans = 0;
    for i in 0..n - 2 {
        for j in i + 1..n - 1 {
            for k in j + 1..n {
                let (a, b, c) = (ls[i], ls[j], ls[k]);
                if a + b > c { ans += counter[&a] * counter[&b] * counter[&c]; }
            }
        }
    }
    println!("{}", ans);
}