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 main(){
	int N,ans=99;
	cin >> N;
	vector<int> D(13,0);
	for(int i=0;i<N;i++){
		int d;
		cin >> d;
		D[d]++;
	}
	int l=0,r=0;
	for(int i=0;i<=12;i++){
		if(i==0){
			if(D[i]>0)ans=0;
		}
		else if(i<12){
			if(D[i]>2)ans=0;
			else if(D[i]==2){
				ans=min(ans,i-r);
				ans=min(ans,i-l);
				l=r=i;
			}
			else if(D[i]==1){
				if(l<r){
					ans=min(ans,i-l);
					l=i;
				}
				else{
					ans=min(ans,i-r);
					r=i;
				}
			}
		}
		else{
			if(D[i]>2)ans=0;
			else if(D[i]==1){
				ans=min(ans,i-max(l,r));
			}
			else ans=min(ans,24-l-r);
		}
	}
	
	cout << ans << endl;
	return 0;
}