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 a[100],tot,cnt[100],b[100],ans,n,x;
void dfs(int x){
	if (x>tot){
		int num=1e9;
		for (int i=1;i<=n+tot;i++)
			for (int j=i+1;j<=n+tot;j++)
				num=min(num,min(24-abs(b[i]-b[j]),abs(b[i]-b[j])));
		ans=max(ans,num);
		return;		
	}
	b[x+n]=a[x];dfs(x+1);
	b[x+n]=24-a[x];dfs(x+1);
}
int main(){
	scanf("%d",&n);
	cnt[0]++;
	while (n--)scanf("%d",&x),cnt[x]++;
	n=0;
	for (int i=0;i<=12;i++)
		if (cnt[i]>2){
			puts("0");
			return 0;
		}
		else if (cnt[i]==2){
			if (!i){
				puts("0");
				return 0;
			}
			else b[++n]=i,b[++n]=24-i;
		}
		else if (cnt[i]==1)a[++tot]=i;
	dfs(1);
	printf("%d\n",ans);
}