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 (2.4.0) (judge verdict: Accepted)

n = int(input())
D = list(map(int, input().split()))
from collections import Counter
C = Counter(D)
#print(C)

A = [0]
C = list(C.items())
C.sort()
#print(C)
for i in range(len(C)):
    k, v = C[i]
    if k != 0 and k != 12:
        if v == 1:
            A.append(k*(-1)**i)
        elif v == 2:
            A.append(k)
            A.append(-k)
        else:
            print(0)
            exit()
    else:
        if v == 1:
            A.append(k*(-1)**i)
        else:
            print(0)
            exit()

#print(A)
m = len(A)
ans = 10**18
for i in range(m-1):
    for j in range(i+1, m):
        temp = min(abs(A[i]-A[j]), 24-abs(A[i]-A[j]))
        ans = min(ans, temp)
print(ans)