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
N = int(input())
D = list(map(int,input().split(" ")))

ans = D[:]
D.sort()
D_1 = []
D_2 = []

for i in range(N):
	if i % 2 == 0:
		D_1.append(24-D[i])
		D_2.append(D[i])
	else:
		D_2.append(24-D[i])
		D_1.append(D[i])


for i in range(N):
	for j in range(i+1,N):
		df1 = abs(D_1[j] - D_1[i])
		df2 = abs(D_2[j] - D_2[i])
		
		df1 = min([df1,24-df1])
		df2 = min([df2,24-df2])
			
		ans.append(df1)
		ans.append(df2)

print(min(ans))




