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 Fortran (gfortran v4.8.4) (judge verdict: Accepted)

program time_gap
  implicit none
  integer :: n, d(50), t(0:12), x, y(11), u, m, s, i, j, k
  logical :: c(0:24), f(0:24)
  d = 0
  t = 0
  y = 0
  c = .false.
  t(0) = 1
  read(*,*) n
  read(*,*) d(1:n)
  do i = 1, n
    t(d(i)) = t(d(i))+1
  end do
  if (t(0).gt.1.or.t(12).gt.1) then
    write(*,'(i0)') 0
    stop
  end if
  x = 0
  c(0) = .true.
  c(24) = .true.
  do i = 1, 11
    if (t(i).gt.2) then
      write(*,'(i0)') 0
      stop
    else if (t(i).eq.2) then
      c(i) = .true.
      c(24-i) = .true.
    else if (t(i).eq.1) then
      x = x+1
      y(x) = i
    end if
  end do
  if (t(12).eq.1) c(12) = .true.
  u = 2**x-1
  m = 0
  do k = 0, u
    f = c
    do i = 1, x
      if (btest(k,i-1)) then
        f(y(i)) = .true.
      else
        f(24-y(i)) = .true.
      end if
    end do
    s = 24
    i = 0
    do
      j = i+1
      do while (.not.f(j))
        j = j+1
      end do
      s = min(s,j-i)
      if (j.eq.24) exit
      i = j
    end do
    m = max(m,s)
  end do
  write(*,'(i0)') m
  stop
end program time_gap