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 ss[13],n,bb[13],tt[5];
inline int cc(int xx)
{
	for(int i=0;i<13;i++) if(ss[i]>=3) return 0;
	int bb=30;
	for(int i=0;i<13;i++) if(ss[i]==2) 
		bb=min(bb,min(2*i%24,24-2*i%24));
	for(int i=0;i<13;i++) if(ss[i])
	{
		for(int j=i+1;j<13;j++) if(ss[j])
		{
			tt[0]=0;
			if(ss[i]==2) tt[++tt[0]]=i,tt[++tt[0]]=-i;
			else tt[++tt[0]]=i*(xx&(1<<i)? -1:1);
			if(ss[j]==2) tt[++tt[0]]=j,tt[++tt[0]]=-j;
			else tt[++tt[0]]=j*(xx&(1<<j)? -1:1);
			for(int t=1;t<=tt[0];t++)
				for(int s=t+1;s<=tt[0];s++)
					bb=min(bb,min(abs(tt[t]-tt[s])%24,24-abs(tt[t]-tt[s])%24));
		}
	}
	return bb;
}
int main()
{
	scanf("%d",&n); ss[0]++;
	for(int i=1;i<=n;i++)
	{
		int k; scanf("%d",&k);
		ss[k]++;
	}
	int ans=0;
	for(int i=0;i<(1<<14);i++)
		ans=max(ans,cc(i));
	cout << ans << endl;
	return 0;
}