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 (1.15.1) (judge verdict: Time Limit Exceeded)

use std::io;
use std::cmp;

fn main() {
    let n = read_values::<usize>()[0];
    let d = read_values::<usize>();

    {
        let mut count = vec![0; 13];
        for e in &d {
            count[*e] += 1;
        }

        if count[0] >= 1 {
            println!("0");
            return;
        }

        for c in &count {
            if *c >= 3 {
                println!("0");
                return;
            }
        }
        if count[12] >= 2 {
            println!("0");
            return;
        }
    }

    let mut max = 0;
    for mask in 0..(1 << n) {
        let mut v = vec![0; n + 1];
        for i in 0..n {
            if ((1 << i) & mask) != 0 {
                v[i] = d[i] as i32;
            } else {
                v[i] = -(d[i] as i32);
            }
        }

        v.sort();
        let mut min = 12;
        for i in 0..(n + 1) {
            if i == n {
                min = cmp::min(min, v[0] + 24 - v[i]);
            } else {
                min = cmp::min(min, v[i + 1] - v[i]);
            }
        }

        max = cmp::max(max, min);
    }
    println!("{}", max);
}

fn read_line() -> String {
    let stdin = io::stdin();
    let mut buf = String::new();
    stdin.read_line(&mut buf).unwrap();
    buf
}

fn read<T>() -> T
    where T: std::str::FromStr,
          T::Err: std::fmt::Debug
{
    read_line().trim().parse().unwrap()
}

fn read_values<T>() -> Vec<T>
    where T: std::str::FromStr,
          T::Err: std::fmt::Debug
{
    read_line()
        .split(' ')
        .map(|a| a.trim().parse().unwrap())
        .collect()
}