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[50],b[50],c[50],ans,x,A[30],t;
void dfs(int x){
	if (x==n+1){
		t=n;
		for (int i=1;i<=n;++i) c[i]=a[i];
		for (int i=0;i<13;++i) if (A[i]==2) c[++t]=-i,c[++t]=i;
		sort(c+1,c+t+1); x=24+c[1]-c[t];
		for (int i=1;i<t;++i) x=min(x,c[i+1]-c[i]);
		ans=max(x,ans); return;
	}
	a[x]=-b[x];
	dfs(x+1);
	a[x]=b[x];
	dfs(x+1);
}
int main(){
	scanf("%d",&n); A[0]=1;
	for (int i=1;i<=n;++i) scanf("%d",&x),++A[x];
	for (int i=0;i<13;++i)
	if (A[i]>2){puts("0"); return 0;}
	n=0;
	for (int i=0;i<13;++i) if (A[i]==1) b[++n]=i;
	dfs(1);
	printf("%d\n",ans);
	return 0;
}