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 Ruby (2.3.3) (judge verdict: Accepted)

N = gets.to_i
D = gets.split.map(&:to_i)

time = Array.new(24, 0)
counter = Hash.new(0)

t1 = []
t2 = []

D.each do |d|
  counter[d] += 1
  t1 << d
  t2 << 24 - d
end

time[0] += 1

one_d = counter.select { |k, v| v == 1 }
two_d = counter.select { |k, v| v == 2 }
three_d = counter.select { |k, v| v >= 3 }

if three_d.size > 0
  puts 0
else
  two_d.each_key do |d|
    time[d] += 1
    time[(24 - d) % 24] += 1
  end

  ans = 0

  0.upto(2 ** 12 - 1) do |mask|
    temp = time.dup

    one_d.each_key do |d|
      if mask[d] == 1
        temp[d] += 1
      else
        temp[(24 - d) % 24] += 1
      end
    end

    next if temp.max >= 2

    d = 0
    min_d = Float::INFINITY

    (1..24).each do |t|
      t %= 24

      if temp[t].zero?
        d += 1
      else
        min_d = d if min_d > d
        d = 0
      end
    end

    if min_d != Float::INFINITY && ans < min_d + 1
      ans = min_d + 1
    end
  end

  puts ans
end