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 "algorithm"
#include "string"
#include "vector"
#include "cmath"
#include "bitset"
#include "queue"
#include "functional"
#include "map"
#include "unordered_map"
#include "set"
#include "stack"
 
#define lp(n) for (int i = 0; i < n; i++)
#define LP(n,i) for (int i = 0; i < n; i++)
#define mod 1000000007
#define sp ' '
#define intmax 2147483647
#define llmax 9223372036854775807
#define nyan "(=^･ω･^=)"
typedef long long ll;
using namespace std;

int n,d[13],mem,mx,mem2;
bool p[24],memp[24];
int main(){
	cin>>n;
	lp(12)
		d[i]=0;
	lp(n){
		cin>>mem;
		if(mem==0){
			cout<<0<<endl;
			return 0;
		}
		d[mem]++;
	}
	if(d[0]>=1||d[12]>=2){
		cout<<0<<endl;
		return 0;
	}
	lp(24)
		p[i]=false;
	p[0]=true;
	if(d[12]==1)
		p[12]=true;
	lp(12){
		if(d[i]>=3){
			cout<<0<<endl;
			return 0;
		}
		if(d[i]==2){
			p[i]=p[24-i]=true;
		}
	}
	mx=0;
	lp((1<<12)-1){
		LP(24,j)
			memp[j]=p[j];
		mem=i;
		LP(12,j){
			if(d[j]==1){
				if(mem%2==1)
					memp[j]=true;
				else
					memp[(24-j)%24]=true;
			}
			mem/=2;
		}
		mem=-10000;
		mem2=100;
		LP(24,j){
			if(memp[j]){
				mem2=min(mem2,j-mem);
				mem=j;
			}
			mem2=min(mem2,24-mem);
		}
		mx=max(mx,mem2);
	}
	cout<<mx<<endl;
	return 0;
}
