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;
typedef long long LL;
typedef pair<LL,LL> PII;


int main() {
	int N;
	cin >> N;
	vector<int> D;
	for(int i=0; i<N; i++) {
		int p;
		cin >> p;
		D.push_back(p);
	}
	
	int zero = 0;
	vector<int> cnt(13, 0);
	for(int i=0; i<D.size(); i++) {
		cnt[D[i]] ++;
	}
	if(cnt[0] == 1) zero=1;
	for(int i=0; i<13; i++)
		if(cnt[i] > 2) zero = 1;
	if(zero) {
		cout << 0 << endl;
		return 0;
	} 
	
	
	int maxim = 0;
	for(int b=0; b<(1<<N); b++) {
		vector<int> res(25, 0);
		
		for(int i=0; i<N; i++) {
			if( (b & (1<<i)) > 0) res[D[i]] += 1;
			else res[24-D[i]] += 1;
		}
		int minim = 24;
		for(int i=0; i<24; i++)
			if(res[i] == 2) minim = 0;
			
		res[0]++;
		res[24]++;
		int prev = 0;
		for(int i=1; i<=24; i++) {
	//		cout << res[i] << " ";
			if(res[i] >= 1) {
				minim  = min(minim, i - prev);
				prev = i;
			}
		}
//		cout << endl;
		maxim = max(maxim, minim);
		
		
	}
	cout<< maxim << endl;
	
	return 0;
}