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[20],c[100],tp,ans;
int chk(){
	int ret=10007;
	for(int i=0;i<=tp;++i)
		for(int j=i+1;j<=tp;++j)
			ret=min(ret,min(abs(c[i]-c[j]),24-abs(c[i]-c[j])));
	return ret;
}
void dfs(int x){
	if(x==13){
		ans=max(ans,chk());
		return ;
	}
	if(a[x]==2)c[++tp]=x,c[++tp]=24-x,dfs(x+1),tp-=2;
	else if(a[x]==1)c[++tp]=x,dfs(x+1),c[tp]=24-x,dfs(x+1),tp--;
	else dfs(x+1);
}
int main(){
	scanf("%d",&n);
	for(int i=1,x;i<=n;++i)
		scanf("%d",&x),a[x]++;
	for(int i=0;i<=12;++i)
		if(a[i]>2)return puts("0"),0;
	dfs(0);
	printf("%d",ans);
}