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.

Examples

Input

3
7 12 8


Output

4


Input

2
11 11


Output

2


Input

1
0


Output

0
from collections import Counter
N = int(input())
ds = list(map(int, input().split()))
c = Counter(ds)
cs = c.most_common()

if N >= 24 or cs[0][1] >= 3 or 0 in c:
  print("0")
  exit()

l = len(cs)
k = 0
s = (1 << 0) + (1 << 24)
while k < l and cs[k][1] == 2:
  s += (1 << cs[k][0]) + (1 << (24 - cs[k][0]))
  k += 1

def process(s, i):
  if i == l:
    return min(map(len, bin(s)[2:].split("1")[1:-1])) + 1
  return max(
    process(s + (1 << cs[i][0]), i+1),
    process(s + (1 << (24 - cs[i][0])), i+1),
  )

print(process(s, k))