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>
using namespace std;

int main() {
	int N;
	cin >> N;
	vector<int> d(13, 0), single;
	++d[0];
	for(int i=0; i<N; ++i){
	    int a;
	    cin >> a;
	    ++d[a];
	}
	int ans=12;
	for(int i=0; i<=12; ++i){
	    if(d[i]>2) ans=0;
	    if(d[i]==0) continue;
	    if(d[i]==1) single.push_back(i);
	    for(int j=0; j<=12; ++j){
	        if(i==j){
	            if(d[i]==2) ans=min(ans, min(i*2, 24-i*2));
	        }
	        else if(d[j]==0) continue;
	        else if(d[i]==2 || d[j]==2) ans=min(ans, abs(i-j));
	        else ans=min(ans, min(i+j, 24-i-j));
	    }
	}
	for(int i=0; i+2<single.size(); ++i) ans=min(ans, single[i+2]-single[i]);
	cout << ans << endl;
	return 0;
}
