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_parens)]
#![allow(unused_imports)]
#![allow(non_upper_case_globals)]
#![allow(non_snake_case)]
#![allow(unused_mut)]
#![allow(unused_variables)]
#![allow(dead_code)]

use itertools::Itertools;
use proconio::input;
use proconio::marker::{Chars, Usize1};

#[allow(unused_macros)]
#[cfg(debug_assertions)]
macro_rules! mydbg {
    //($arg:expr) => (dbg!($arg))
    //($arg:expr) => (println!("{:?}",$arg));
      ($($a:expr),*) => {
          eprintln!(concat!($(stringify!($a), " = {:?}, "),*), $($a),*);
      }
}
#[cfg(not(debug_assertions))]
macro_rules! mydbg {
    ($($arg:expr),*) => {};
}

macro_rules! echo {
    ($($a:expr),*) => {
          $(println!("{}",$a))*
      }
}

use std::cmp::*;
use std::collections::*;
use std::ops::{Add, Div, Mul, Sub};

#[allow(dead_code)]
static INF_I64: i64 = 92233720368547758;
#[allow(dead_code)]
static INF_I32: i32 = 21474836;
#[allow(dead_code)]
static INF_USIZE: usize = 18446744073709551;
#[allow(dead_code)]
static M_O_D: usize = 1000000007;
#[allow(dead_code)]
static PAI: f64 = 3.1415926535897932;

trait IteratorExt: Iterator {
    fn toVec(self) -> Vec<Self::Item>;
}

impl<T: Iterator> IteratorExt for T {
    fn toVec(self) -> Vec<Self::Item> {
        self.collect()
    }
}

trait CharExt {
    fn toNum(&self) -> usize;
    fn toAlphabetIndex(&self) -> usize;
    fn toNumIndex(&self) -> usize;
}
impl CharExt for char {
    fn toNum(&self) -> usize {
        return *self as usize;
    }
    fn toAlphabetIndex(&self) -> usize {
        return self.toNum() - 'a' as usize;
    }
    fn toNumIndex(&self) -> usize {
        return self.toNum() - '0' as usize;
    }
}

trait VectorExt {
    fn joinToString(&self, s: &str) -> String;
}
impl<T: ToString> VectorExt for Vec<T> {
    fn joinToString(&self, s: &str) -> String {
        return self
            .iter()
            .map(|x| x.to_string())
            .collect::<Vec<_>>()
            .join(s);
    }
}

trait StringExt {
    fn get_reverse(&self) -> String;
}
impl StringExt for String {
    fn get_reverse(&self) -> String {
        self.chars().rev().collect::<String>()
    }
}

trait UsizeExt {
    fn pow(&self, n: usize) -> usize;
}
impl UsizeExt for usize {
    fn pow(&self, n: usize) -> usize {
        return ((*self as u64).pow(n as u32)) as usize;
    }
}

fn main() {
    input! {
       N: usize,
       mut L:[usize;N],
    }

    L.sort();
    let mut ans: usize = 0;
    for i in 0..N {
        for j in i + 1..N {
            for k in j + 1..N {
                if (L[i] != L[j] && L[j] != L[k] && L[i] != L[k]) && (L[i] + L[j] > L[k]) {
                    mydbg!(L[i], L[j], L[k]);
                    ans += 1;
                }
            }
        }
    }

    echo!(ans);
}