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
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <vector>
#include <algorithm>
#include <cstdlib>
#include <cmath>
#include <map>
#include <set>
#include <string>
#include <iostream>
#include <cassert>
 
#pragma comment(linker, "/STACK:336777216")  
 
using namespace std;

const int MAXN = 200000 + 10;
const int INF = 1000000000;

int N;
int cnt[13];
int used[24];
int ans;

void DFS(int i)
{
	if (i == 13)
	{
		int cur = 100;
		for (int a = 0; a < 24; a ++)
		{
			if (used[a])
			{
				for (int b = 1; b <= 24; b ++)
					if (used[(a + b) % 24] || used[(a - b + 24) % 24])
					{
						cur = min(cur, b);
						break;
					}
			}
		}
		if (cur > ans) ans = cur;
		return;
	}
	if (cnt[i] == 0)
		DFS(i + 1);
	else if (cnt[i] == 2)
	{
		used[i] = used[24 - i] = 1;
		DFS(i + 1);
		used[i] = used[24 - i] = 0;
	}
	else
	{
		used[i] = 1;
		DFS(i + 1);
		used[i] = 0;
		if (i != 0)
		{
			used[24 - i] = 1;
			DFS(i + 1);
			used[24 - i] = 0;
		}
	}
}

void Work()
{
	scanf("%d", &N);
	cnt[0] = 1;
	for (int i = 0; i < N; i ++)
	{
		int d;
		scanf("%d", &d);
		cnt[d] ++;
	}
	if (cnt[0] > 1 || cnt[12] > 1)
	{
		printf("0\n");
		return;
	}
	for (int i = 1; i <= 11; i ++)
		if (cnt[i] > 2)
		{
			printf("0\n");
			return;
		}
	ans = 0;
	DFS(0);
	printf("%d\n", ans);
}

int main()
{
	Work();
	return 0;
}
