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 Python (3.4.3) (judge verdict: Accepted)

import sys;
input = sys.stdin.readline

n = int(input())
d = [int(x) for x in input().split()]

def doit():
  if n >= 24:
    return 0
  if n == 1:
    return d[0]
  
  res = [0] * 13
  for x in d:
    if x == 0:
      return 0
    res[x] += 1;
    if x == 12:
      if res[x] == 2:
        return 0
    if res[x] == 3:
      return 0
  
  ban = [False] * 24
  mm = 12
  flip = False
  prev = [0, 24]
  for i, x in enumerate(res):
    diff = 12
    if x == 2:
      ban[i] = ban[24 - i] = True
      diff = i - prev[0]
      prev[0] = i
      prev[1] = 24 - i
      print("x2", i, diff, file=sys.stderr)
      flip = not flip
    if x == 1:
      if flip:
        ban[i] = True
        diff = i - prev[0]
        prev[0] = i
      else:
        ban[24 - i] = True
        diff = prev[1] - (24 - i)
        prev[1] = 24 - i
      flip = not flip
      print("x1", i, diff, file=sys.stderr)
    if diff < mm:
      mm = diff
      print(mm, file=sys.stderr)
  #last = abs(prev[1] - prev[0])
  last = prev[1] - prev[0]
  print("last", prev[0], prev[1], file=sys.stderr)
  if last < mm:
    return last
  return mm


print(doit())