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 <vector>
#include <algorithm>

using namespace std;
int main(){
	int i,j,n,d[30],s = 0,c[30]={0};
	cin >> n;
	if(n>23){
		cout << 0 << endl;
		return 0;
	}
	
	for(i=0;i<n;i++){
		cin >> d[i];
		c[d[i]]++;
		if(!d[i]){
			cout << 0 << endl;
			return 0;
		}
	}
	
	for(i=1;i<=11;i++){
		if(c[i]>=3 || c[12]>=2){
			cout << 0 << endl;
			return 0;
		}
	}
	
	if(n==23){
		cout << 1 << endl;
		return 0;
	}
	
	for(i=0;i<(1<<n);i++){
		vector<int> v;
		for(j=0;j<n;j++){
			if(i & 1<<j){
				v.push_back((d[j]+12)%24);
			}else{
				v.push_back(12-d[j]);
			}
		}
		v.push_back(12);
		sort(v.begin(),v.end());
		int x = (24 + v[0] - v.back())%24;
		for(j=0;j<v.size()-1;j++){
			x = min(x,v[j+1]-v[j]);
		}
		s = max(s,x);
	}
	cout << s << endl;
}