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 solve(const vector<int>& times){
	if(times[0] >= 2 || times[12] >= 2) return 0;
	for(int i=1;i<12;i++) if(times[i] >= 3) return 0;
	int res = 0;
	for(int i=0;i<(1<<11);i++){
		vector<int> exist(25, 0);
		if(times[0] > 0) exist[0] = exist[24] = 1;
		if(times[12] > 0) exist[12] = 1;
		for(int j=1;j<12;j++){
			if(times[j] == 0) continue;
			if(times[j] == 1) exist[(i&(1<<(j-1))) ? j : 24-j] = 1;
			if(times[j] == 2) exist[j] = exist[24-j] = 1;
		}
		int pre = 0;
		int cur_res = 100;
		for(int j=1;j<=24;j++){
			if(exist[j]){
				cur_res = min(cur_res, j-pre);
				pre = j;
			}
		}
		res = max(res, cur_res);		
	}
	return res;
}

int main(){
	int N;
	while(cin >> N){
		vector<int> cnt(13, 0);
		cnt[0] = 1;
		for(int i=0;i<N;i++){
			int t; cin >> t;
			cnt[t]++;			
		}
		cout << solve(cnt) << endl;
	}
}
