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

N = int(sys.stdin.readline().strip())
D = [0] + [int(x) for x in sys.stdin.readline().strip().split()]

def kei(xs, n):
    xs = xs + [x + 24 for x in xs]
    m1 = min(abs(n - x) for x in xs)
    m2 = min(abs((24 - n) - x) for x in xs)
    if m1 >= m2:
        return n, m1
    else:
        return 24 - n, m2

def calc(D):
    D.sort()
    #print(D)
    n = len(D)
    if N > 24:
        return 0
    if D[1] == 0:
        return 0
    xs = D[:2]
    D = D[2:]
    r = xs[1] - xs[0]
    for d in D:
        #print(xs, d)
        c, m = kei(xs, d)
        #print("r", c, m)
        if m < r:
            r = m
        xs += [c]
    return r



print(calc(D))
