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 PyPy3 (7.3.0) (judge verdict: Accepted)

n = int(input())
dl = list(map(int, input().split()))

if 0 in dl:
    print(0)
    exit()

dl.sort()
dl1 = [0,24]
dl2 = [0,24]
for i, d in enumerate(dl):
    if i%2 == 0:
        dl1.append(d)
        dl2.append(24-d)
    else:
        dl1.append(24-d)
        dl2.append(d)

dl1.sort()
dl2.sort()

min_1 = 24
for a,b in zip(dl1[1:],dl1[:-1]):
    diff = a-b
    min_1 = min(min_1,diff)

min_2 = 24
for a,b in zip(dl2[1:],dl2[:-1]):
    diff = a-b
    min_2 = min(min_2,diff)

ans = max(min_1,min_2)
print(ans)