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::io::*;
use std::str::*;

fn read<T: FromStr>(s: &mut StdinLock) -> Option<T> {
    let s = s.by_ref().bytes().map(|c| c.unwrap() as char)
        .skip_while(|c| c.is_whitespace())
        .take_while(|c| !c.is_whitespace())
        .collect::<String>();
    s.parse::<T>().ok()
}

fn main(){
    let s = stdin();
    let mut s = s.lock();
    let s = &mut s;

    let n = read::<usize>(s).unwrap();
	let mut v : Vec<i32> = Vec::with_capacity(n );
	for _ in 0 .. n {
		v.push(read::<i32>(s).unwrap());
	}
	let mut cnt = 0;
	for i  in 0 .. n {
		if i+1 == n {
			continue;
		} 
		for j in i+1 .. n {
			if j+1 == n {
				continue;
			}
			for k  in j+1 .. n {
				if v[i] == v[j] || v[i] == v[k] || v[j] == v[k] {
					continue;
				}
				if v[i] + v[j] > v[k] 
				&& v[i] + v[k] > v[j]
				&& v[k] + v[j] > v[i] {
					cnt += 1;
				}	
			}
		}
	}
	println!("{}", cnt);
}