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 <climits>
#include <stack>
#include <queue>
#include <string>
#include <set>
#include <map>
#include <math.h>
#include <algorithm>
#include <vector>
 
using namespace std;
 
typedef long long ll;
 
typedef pair<ll,ll> P;
long long int INF = 1e18;
double Pi = 3.141592653589;
const int mod = 1000000007;
// memset(a,0,sizeof(a)); →全部０にする
 
vector<int> G[100005];
std::vector<P> tree[100010];
 
int dx[8]={1,0,-1,0,1,1,-1,-1};
int dy[8]={0,1,0,-1,1,-1,-1,1};
 
#define p(x) cout<<x<<endl;
#define e cout<<endl;
#define pe(x) cout<<x<<" ";
#define re(i,a,b) for(i=a;i<=b;i++)
 
ll i,j,k,l,ii,jj;
ll n,m;
ll x,y;
ll a[55],b[25];
ll ans=0;
ll kouho=INF;
 
int main(){
	cin>>n;
	if(n>=14){
		ans=1;
		for(i=0;i<n;i++){
			cin>>a[i];
		}
		for(i=0;i<=12;i++){
			b[i]=0;
		}
		for(i=0;i<n;i++){
			m=a[i];
			b[m]++;
		}
		for(i=1;i<=11;i++){
			if(b[i]>=3){
				ans=0;
			}
		}
		if(b[12]>=2 || b[0]>=1)ans=0;
		//0ninarubeki nanoni 0zyanai

	}else{
		for(i=0;i<n;i++){
			cin>>a[i];
		}
		a[n]=0;
		for(i=0;i<=16383;i++){
			for(j=0;j<n+1;j++){
				if(i & 1<<j){
					b[j]=a[j];
				}else{
					b[j]=24-a[j];
				}
			}
			kouho=INF;
			for(k=0;k<n+1;k++){
				for(j=0;j<n+1;j++){
					if(k==j)continue;
					kouho=min(min(24-abs(b[k]-b[j]),abs(b[k]-b[j])),kouho);
				}
			}
		ans=max(ans,kouho);
		}
	}
	if(n==1)ans=min(a[0],24-a[0]);
	p(ans);
				
	return 0;
}