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<cstdio>
#include<vector>
#include<algorithm>
using namespace std;

int N;
int num[13];

int main(){
	scanf("%d", &N);
	for(int i = 0; i < N; i++){
		int D;
		scanf("%d", &D);
		num[D]++;
	}
	num[0]++;
	vector<int> va, vb;
	int cnt = 0;
	for(int i = 0; i < 13; i++){
		if(i == 0 || i == 12){
			if(num[i] >= 2){
				printf("0\n");
				return 0;
			}
		}else{
			if(num[i] >= 3){
				printf("0\n");
				return 0;
			}
		}
		if(num[i] == 1){
			va.push_back(i);
			cnt++;
		}else if(num[i] == 2){
			vb.push_back(i);
			vb.push_back(24 - i);
		}
	}
	for(int i = 0; i < vb.size(); i++) va.push_back(vb[i]);
	int res = 0;
	for(int i = 0; i < (1 << cnt); i++){
		int min_diff = 13;
		for(int j = 0; j < va.size(); j++){
			int time_j;
			if(j < cnt) time_j = (i & (1 << j) ? va[j] : ((24 - va[j]) % 24));
			else time_j = va[j];
			for(int k = 0; k < va.size(); k++){
				if(k == j) continue;
				int time_k;
				if(k < cnt) time_k = (i & (1 << k) ? va[k] : ((24 - va[k]) % 24));
				else time_k = va[k];
				int time_diff = abs(time_j - time_k);
				min_diff = min(min_diff, min(time_diff, 24 - time_diff));
			}
		}
		res = max(res, min_diff);
	}
	printf("%d\n", res);
	return 0;
}
