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
import sys
input = sys.stdin.readline
sys.setrecursionlimit(10 ** 7)

import itertools

N = int(input())
D = [int(x) for x in input().split()]

counter = [1] + [0] * 12
for x in D:
    counter[x] += 1

bl = counter[0] > 1 or counter[12] > 1 or any(x >= 3 for x in counter)
if bl:
    print(0)
    exit()

itrs = [[[0]]]
for x in range(1,12):
    if counter[x] == 0:
        itr = [[]]
    elif counter[x] == 1:
        itr = [[x],[-x]]
    else:
        itr = [[x,-x]]
    itrs.append(itr)
itrs.append([[]] if counter[12] == 0 else [[12]])

def F(arr):
    f = lambda x,y: min((x-y)%24, (y-x)%24)
    return min(f(x,y) for x,y in itertools.combinations(arr,2))

answer = 0
for p in itertools.product(*itrs):
    arr = []
    for x in p:
        arr += x
    answer = max(answer, F(arr))

print(answer)