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,*D = `dd`.split.map &:to_i

nums = Hash[(0..12).map{ |d| [d,0] }]
nums[0] = 1
D.each{ |d| nums[d] += 1 }
if nums[0] >= 2 || nums[12] >= 12 || (1..11).any?{ |d| nums[d] >= 3 }
  p 0; exit
end

def dig(d, nums)
  return [1 + (1 << 24)] if d == 13
  
  v = dig(d+1,nums)
  n = nums[d]
  v.map{ |x|
    a = (1 << d); b = (1 << (24-d))
    case n
    when 2; [x | a | b]
    when 1; [x | a, x | b]
    else    [x]
    end
  }.flatten
end

p dig(0, nums).map{ |n|
  n.to_s(2).scan(/(?<=1)0*(?=1)/).map{ |d| d.size+1 }.min || 0
}.max