Problem:
Score : 500 points

Problem Statement

In CODE FESTIVAL XXXX, there are N+1 participants from all over the world, including Takahashi.

Takahashi checked and found that the time gap (defined below) between the local times in his city and the i-th person's city was D_i hours.
The time gap between two cities is defined as follows. For two cities A and B, if the local time in city B is d o'clock at the moment when the local time in city A is 0 o'clock, then the time gap between these two cities is defined to be min(d,24-d) hours.
Here, we are using 24-hour notation.
That is, the local time in the i-th person's city is either d o'clock or 24-d o'clock at the moment when the local time in Takahashi's city is 0 o'clock, for example.

Then, for each pair of two people chosen from the N+1 people, he wrote out the time gap between their cities. Let the smallest time gap among them be s hours.

Find the maximum possible value of s.

Constraints

1 \leq N \leq 50

0 \leq D_i \leq 12

All input values are integers.

Input

Input is given from Standard Input in the following format:

N
D_1 D_2 ... D_N

Output

Print the maximum possible value of s.

Sample Input 1

3
7 12 8

Sample Output 1

4

For example, consider the situation where it is 7, 12 and 16 o'clock in each person's city at the moment when it is 0 o'clock in Takahashi's city. In this case, the time gap between the second and third persons' cities is 4 hours.

Sample Input 2

2
11 11

Sample Output 2

2

Sample Input 3

1
0

Sample Output 3

0

Note that Takahashi himself is also a participant.

Solution in Rust (1.42.0) (judge verdict: Wrong Answer)

use proconio::input;
use std::collections::HashMap;

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

    let mut map = HashMap::new();
    for d in dd {
        *map.entry(d).or_insert(0) += 1;
    }

    let mut ans = 0;

    for i in 0..1<<map.len() {
        let mut times = vec![];
        for (j, (&key, &val)) in map.iter().enumerate() {
            if val > 2 {
                println!("0");
                return;
            } else if val == 2 {
                times.push(key);
                times.push(24-key);
            } else {
                if i & 1<<j == i<<j {
                    times.push(24-key);
                } else {
                    times.push(key);
                }
            }
        }
        ans = std::cmp::max(ans, min_time(times));
    }
    
    println!("{}", ans);
}

fn min_time(mut times: Vec<usize>) -> usize {
    times.sort();
    
    let mut min = 12;

    for i in 0..times.len()-1 {
        let sub = times[i+1] - times[i];
        min = std::cmp::min(min, sub);
    }

    min = std::cmp::min(min, times[0]);
    min = std::cmp::min(min, 24 - times[times.len()-1]);
    min
}