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: Wrong Answer)

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

from collections import defaultdict
d = defaultdict(int)


for d_ in ds:
    d[d_] += 1
    if d_ == 12 and d[d_] == 2:
        print(0)
        exit()
    elif d_ == 0 and d[d_] == 2: 
        print(0)
        exit()
    elif d[d_] == 3:
        print(0)
        exit()

def calcmin(tmp):
    n = len(tmp)
    ret = 1 << 50
    for i in range(n):
        for j in range(i+1,n):
            ret = min(ret, abs(tmp[i]-tmp[j]))
    return ret

two = []
x = []
for k in d.keys():
    if d[k] == 2 or k == 0 or k == 12:
        two.append(k)
        two.append(24-k)
    elif d[k] == 1:
        x.append(k)
two = list(set(two))
n = len(x)

ans = 0
for i in range(1 << n):
    tmp = x[:]
    for j in range(n):
        if i & (1 << j):
            tmp[j] = 24-x[j]
    tmp.extend(two)
    ans = max(calcmin(tmp), ans)
print(ans)