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<iostream>
#include<algorithm>
using namespace std;
int n,c[13]={1};
int main()
{
	cin>>n;
	for(int i=0;i<n;i++)
	{
		int a;cin>>a;
		c[a]++;
	}
	int ans=0;
	for(int i=1;i<12;i++)if(c[0]>1||c[12]>1||c[i]>2){
		cout<<0<<endl;
		return 0;
	}
	for(int i=0;i<1<<11;i++)
	{
		int d[30]={};
		for(int j=0;j<=12;j++)
		{
			if(c[j]==2)d[j]=d[24-j]=1;
			else if(c[j]==1)
			{
				if(0<j&&j<12)d[i&(1<<(j-1))?j:24-j]=1;
				else d[j]=1;
			}
		}
		int now=99,cnt=1;
		for(int j=1;j<24;j++)
		{
			if(d[j])now=min(now,cnt),cnt=1;
			else cnt++;
		}
		ans=max(ans,min(now,cnt));
	}
	cout<<ans<<endl;
}