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 <string>
#include <vector>
#include <algorithm>
#include <cstdlib>

using namespace std;

int main(){
	ios_base::sync_with_stdio(false);
	int n;
	cin >> n;
	vector<int> h(13);
	++h[0];
	for(int i = 0; i < n; ++i){
		int di;
		cin >> di;
		++h[di];
	}
	if(h[0] >= 2 || h[12] >= 2){
		cout << 0 << endl;
		return 0;
	}
	for(int i = 1; i < 12; ++i){
		if(h[i] >= 3){
			cout << 0 << endl;
			return 0;
		}
	}
	int fixed = 0, variable = 0;
	if(h[0]  == 1){ fixed |= (1 << 12); }
	if(h[12] == 1){ fixed |= (1 <<  0); }
	for(int i = 1; i < 12; ++i){
		if(h[i] == 2){
			fixed |= (1 << (12 - i));
			fixed |= (1 << (12 + i));
		}else if(h[i] == 1){
			variable |= (1 << i);
		}
	}
	int answer = 0;
	for(int i = ((1 << 13) - 1) & variable; i >= 0; i = ((i - 1) & variable)){
		int set = fixed;
		for(int j = 0; j <= 12; ++j){
			if(variable & (1 << j)){
				if(i & (1 << j)){
					set |= (1 << (12 + j));
				}else{
					set |= (1 << (12 - j));
				}
			}
		}
		int s = 24;
		int last = __builtin_ctz(set);
		for(int j = last + 1; j < 24; ++j){
			if(set & (1 << j)){
				const int t = j - last;
				s = min(s, min(t, 24 - t));
				last = j;
			}
		}
		const int t = last - __builtin_ctz(set);
		if(t > 0){ s = min(s, min(t, 24 - t)); }
		answer = max(answer, s);
		if(i == 0){ break; }
	}
	cout << answer << endl;
	return 0;
}

