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: Time Limit Exceeded)

// #![allow(unused_mut)]
#![allow(dead_code)]
#![allow(unused_macros)]

use proconio::{input};

// The debug version
#[cfg(feature = "debug")]
macro_rules! debug {
    ($( $args:expr ),*) => { eprintln!( $( $args ),* ); }
}
// Non-debug version
#[cfg(not(feature = "debug"))]
macro_rules! debug {
       ($( $args:expr ),*) => {}
}


use num_traits::int::PrimInt;
// #[overflow_checks(off)]
// i32 -> i32, i64 -> i64
fn accum<T: PrimInt>(a: &Vec<T>) -> Vec<T> {
    let n: usize = a.len();
    let mut ac: Vec<T> = Vec::with_capacity(n + 1);
    ac.push(a[0] - a[0]);  // ac.push(0) がしたいんですけど
    for i in 0..n {
        ac.push(ac[i] + a[i]);
    }
    ac
}
/// i32 -> i64
fn accum64(a: &Vec<i32>) -> Vec<i64> {
    let n: usize = a.len();
    let mut ac: Vec<i64> = Vec::with_capacity(n + 1);
    ac.push(0);
    for i in 0..n {
        ac.push(ac[i] + i64::from(a[i]));
    }
    ac
}

fn rle<T: PrimInt>(a: &Vec<T>) -> Vec<(T,u32)> {
    let mut res: Vec<(T,u32)> = Vec::with_capacity(0);

    let mut last : T = T::zero();
    let mut cnt : u32 = 0;

    for c in a.iter() {
        if *c == last {
            cnt += 1;
        } else {
            if cnt > 0 {
                res.push((last, cnt));
            }
            last = *c;
            cnt = 1;
        }
    };
    if cnt > 0 {
        res.push((last, cnt));
    }
    res
}
fn rle_str(a: &str) -> Vec<(char,u32)> {
    let mut res: Vec<(char,u32)> = Vec::with_capacity(0);

    let mut last : char = '\0';
    let mut cnt : u32 = 0;

    for c in a.chars() {
        if c == last {
            cnt += 1;
        } else {
            if cnt > 0 {
                res.push((last, cnt));
            }
            last = c;
            cnt = 1;
        }
    };
    if cnt > 0 {
        res.push((last, cnt));
    }
    res
}

fn yesno(b: bool) -> &'static str {
    if b {
        return "Yes";
    } else {
        return "No";
    }
}

fn solve(n: usize, mut a: Vec<i32>) -> u32 {
    let mut cnt: u32 = 0;

    a.sort();

    for i in 0..(n-2) {
        for j in i+1..n-1 {
            for k in j+1..n {
                // ai <= aj <= ak
                if a[i] == a[j] || a[j] == a[k] || a[k] == a[i] {
                    continue;
                }
                if a[i] + a[j] > a[k] {
                    // println!("{} {} {}", a[i], a[j], a[k]);
                    cnt += 1;
                }
            }
        }
    }
    cnt
}

fn main() {
    input! {
        n: usize,
        a: [i32; n],
    }
    debug!("Input: n={}, a={:?}", n, a);

    let ans: u32 = solve(n, a);
    println!("{}", ans);
}