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<cstring>
using namespace std;

int main(){
	int n;
	int i,j,k;
	int d[15];
	int s=12;
	memset(d,0,sizeof(d));
	cin>>n;
	for(i=0;i<n;i++){
		cin>>k;
		d[k]++;
	}
	int a=0,b=0;
	for(i=0,j=0;i<=12;i++){
		if(d[i]==0)continue;
		if(d[i]>2)s=0;
		else if(d[i]==2){
			if(s>i-b)s=i-b;
			b=i;
			if(s>a+i)s=a+i;
			a=-i;
		}else if(j%2==0){
			if(s>i-b)s=i-b;
			b=i;
		}else{
			if(s>a+i)s=a+i;
			a=-i;
		}
		j++;
	}
	a+=24;
	if(s>a-b)s=a-b;
	cout<<s<<endl;
	return 0;
}
