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())
D = list(map(int,input().split()))
kouho_1 = [0]*(n+1)
kouho_2 = [0]*(n+1)
flag = False
change = 0
for i in range(n):
    if D[i] == 0:
        kouho_1[i+1] = 0
    elif D[i] == 12:
        kouho_1[i+1] = 12
    else:
        if not flag:
            flag = True
            kouho_1[i+1] = D[i]
        else:
            cur1 = 24
            temp1 = D[i]
            cur2 = 24
            temp2 = 24-D[i]
            for j in range(i+1):
                cur1 = min(cur1, abs(kouho_1[j]-temp1))
            for j in range(i+1):
                cur2 = min(cur2, abs(kouho_1[j]-temp2))
            if cur1 > cur2:
                kouho_1[i+1] = temp1
            else:
                kouho_1[i+1] = temp2
flag = False
for i in range(n):
    if D[i] == 0:
        kouho_2[i+1] = 0
    elif D[i] == 12:
        kouho_2[i+1] = 12
    else:
        if not flag:
            flag = True
            kouho_2[i+1] = 24-D[i]
        else:
            cur1 = 24
            temp1 = D[i]
            cur2 = 24
            temp2 = 24-D[i]
            for j in range(i+1):
                cur1 = min(cur1, abs(kouho_2[j]-temp1))
            for j in range(i+1):
                cur2 = min(cur2, abs(kouho_2[j]-temp2))
            if cur1 > cur2:
                kouho_2[i+1] = temp1
            else:
                kouho_2[i+1] = temp2
ans1 = 24
ans2 = 24
for i in range(n+1):
    for j in range(i+1, n+1):
        ans1 = min(ans1, abs(kouho_1[i]-kouho_1[j]))
        ans2 = min(ans2, abs(kouho_2[i]-kouho_2[j]))
print(max(ans1,ans2))