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
def inpl(): return [int(i) for i in input().split()]
import sys
N = int(input())

D = inpl()
tt = [0 for _ in range(24)]
tt[0] = 2
for d in D:
    tt[d] += 1
    tt[-d] += 1    

for t in tt:
    if t > 2:
        print(0)
        sys.exit()
if N > 11:
    prev = tt[-1]
    for t in tt:
        if t == 0:
            prev = 0
            continue
        if t == 1:
            if prev == 2:
                print(1)
                break
            prev = 1
        if t == 2:
            if prev == 1 or prev == 2:
                print(1)
                break
            prev = 2
    else:
        print(2)
    sys.exit()
ans = 0
for i in range(2**N):
    stri = format(i, 'b')
    stri = '0'*(N-len(stri)) +stri
    tt = [0 for _ in range(24)]
    tt[0] = 1
    for j in range(N):
        c = 1
        if stri[j] == '0':
             c = -1
        tt[c * D[j]] += 1
    for t in tt:
        if t > 1:
            break
    else:
        ctr = 1
        H = []
        for i in range(48):
            if tt[i%24] == 0:
                ctr += 1
            else:
                H.append(ctr)
                ctr = 1
        H.pop(0)
        ans = max(ans, min(H))
print(ans)