output_description: Print the maximum possible value of s.
  
  * * *
input_description: Input is given from Standard Input in the following format:
  
      
      
      N
      D_1 D_2 ... D_N
attempt: import sys, collections as cl, bisect as bs
  
  sys.setrecursionlimit(100000)
  input = sys.stdin.readline
  mod = 10**9 + 7
  Max = sys.maxsize
  
  
  def l():  # intのlist
      return list(map(int, input().split()))
  
  
  def m():  # 複数文字
      return map(int, input().split())
  
  
  def onem():  # Nとかの取得
      return int(input())
  
  
  def s(x):  # 圧縮
      a = []
      aa = x[0]
      su = 1
      for i in range(len(x) - 1):
          if aa != x[i + 1]:
              a.append([aa, su])
              aa = x[i + 1]
              su = 1
          else:
              su += 1
      a.append([aa, su])
      return a
  
  
  def jo(x):  # listをスペースごとに分ける
      return " ".join(map(str, x))
  
  
  def max2(x):  # 他のときもどうように作成可能
      return max(map(max, x))
  
  
  def In(x, a):  # aがリスト(sorted)
      k = bs.bisect_left(a, x)
      if k != len(a) and a[k] == x:
          return True
      else:
          return False
  
  
  """
  def nibu(x,n,r):
      ll = 0
      rr = r
      while True:
          mid = (ll+rr)//2
  
      if rr == mid:
          return ll
      if (ここに評価入れる):
          rr = mid
      else:
          ll = mid+1
  """
  
  
  n = onem()
  
  d = l()
  
  d.sort()
  
  a = [0 for i in range(24)]
  a[0] += 1
  
  for i in range(n):
      if i % 2 == 0:
          a[d[i]] += 1
      else:
          if d[i]:
              a[24 - d[i]] += 1
          else:
              a[0] += 1
  
  co = 12
  
  for i in range(24):
      if a[i] >= 2:
          co = 0
          break
      for j in range(i + 1, 24):
          if a[i] >= 1 and a[j] >= 1:
              co = min(co, min(24 - (j - i), j - i))
  print(co)
  
problem_description: 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.
samples: [{"input": "3\n    7 12 8", "output": "4\n    \n\nFor example, consider the situation where it is 7, 12 and 16 o'clock in each\nperson's city at the moment when it is 0 o'clock in Takahashi's city. In this\ncase, the time gap between the second and third persons' cities is 4 hours.\n\n* * *"}, {"input": "2\n    11 11", "output": "2\n    \n\n* * *"}, {"input": "1\n    0", "output": "0\n    \n\nNote that Takahashi himself is also a participant."}]