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;

#define REP(i,n) for(int i=0;i<n;i++)
#define REPR(i,n) for(int i=n;i>=0;i--)
#define FOR(i,m,n) for(int i=m;i<n;i++)
#define EPS (1e-7)
#define INF 1e18

typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
typedef pair<double,double> pdd;

int vx[]={0,1,0,-1};
int vy[]={1,0,-1,0};

int dx[]={1,0,-1,1,-1,1,0,-1};
int dy[]={1,1,1,0,0,-1,-1,-1};

const ll mod=1e9+7;
const int MAX_N=300;

int main(){
	int n;
	int ans=0;
	cin>>n;
	int d[n+1];
	int min_id=0;
	d[0]=0;
	for(int i=1;i<n+1;i++){
		cin>>d[i];
	}
	if(n>=24)ans=0;
	else if(n==23){
		int hours[13];
		fill(hours,hours+13,0);
		for(int i=0;i<n+1;i++){
			hours[d[i]]++;
		}
		bool judge=true;
		for(int i=0;i<13;i++){
			if(i==0 || i==12){
				if(hours[i]>1)judge=false;
			}else{
				if(hours[i]>2)judge=false;
			}
		}
		if(judge)ans=1;
	}else{
		for(int i=0;i<(1<<n);i++){
			int min_time=12;
			int time[n+1];
			time[0]=0;
			for(int j=0;j<n;j++){
				if(((i>>j)&1)==1)time[j+1]=24-d[j+1];
				else time[j+1]=d[j+1];
			}
			for(int j=0;j<n+1;j++){
				for(int k=j+1;k<n+1;k++){
					min_time=min(min_time,min(24-abs(time[j]-time[k]),abs(time[j]-time[k])));
				}
			}
			ans=max(ans,min_time);
		}
	}
	cout<<ans<<endl;
	return 0;
}