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 <bits/stdc++.h>
using namespace std;
int n,a[57],cnt[57],b[57],now=0;
void dfs(int pos)
{
	if (pos>n)
	{
		for (int i=1;i<=n;i++)
			b[i]=a[i];
		sort(b+1,b+n+1);
		int res=30;
		for (int i=1;i<n;i++)
			res=min(res,b[i+1]-b[i]);
		res=min(res,min(b[n]-b[1],24-b[n]+b[1]));
		now=max(now,res);
	} else
	{
		if (a[pos]!=0&&a[pos]!=12)
		{
			a[pos]=24-a[pos];
			dfs(pos+1);
			a[pos]=24-a[pos];
		}
		dfs(pos+1);
	}
}
int main()
{
	scanf("%d",&n);
	for (int i=1;i<=n;i++)
		scanf("%d",&a[i]);
	memset(cnt,0,sizeof(cnt));
	++cnt[0];
	for (int i=1;i<=n;i++)
		++cnt[a[i]];
	++n;
	a[n]=0;
	for (int i=0;i<=12;i++)
	{
		if ((i==0||i==12)&&cnt[i]>1)
		{
			cout << 0 << endl;
			return 0;
		}
		else if (cnt[i]>2)
		{
			cout << 0 << endl;
			return 0;
		}
	}
	dfs(1);
	printf("%d\n",now);
	return 0;
}
