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: Wrong Answer)

fn main() {
    let stdin = std::io::stdin();
    let mut buf = String::new();
    stdin.read_line(&mut buf).unwrap();
    buf.clear();
    stdin.read_line(&mut buf).unwrap();
    let mut ds: Vec<i32> = buf.split_whitespace().map(|s| s.parse().unwrap()).collect();
    ds.push(0);
    ds.sort();
    let mut ps = vec![];
    for i in 0..ds.len() {
        if i % 2 == 1 {
            ds[i] = (24 - ds[i]) % 24;
        }
        if 0 < i {
            let d = ds[i] - ds[i - 1];
            ps.push(d.abs());
        }
    }
    println!("{}", ps.iter().min().unwrap());
}