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;

int main(){
	int n,cnt[13] = {},ma = 0;
	cin >> n;
	for(int i = 0;i < n;i++){
		int d;
		cin >> d;
		cnt[d]++;
	}
	cnt[0]++;
	for(int i = 0;i <= 12;i++){
		if(cnt[i] >= 3){
			cout << 0 << endl;
			return 0;
		}
	}
	for(int i = 0;i < (1 << 13);i++){
		int mi = 30;
		vector<int> vec;
		for(int j = 0;j < 13;j++){
			if(!cnt[j]) continue;
			if(cnt[j] == 2 || ((i >> j) & 1)) vec.push_back(j);
			if(cnt[j] == 2 || !((i >> j) & 1)) vec.push_back(24 - j);
		}
		for(int j = 0;j < vec.size();j++){
			for(int k = j + 1;k < vec.size();k++) mi = min(mi,min(abs(vec[j] - vec[k]),24 - abs(vec[j] - vec[k])));
		}
		ma = max(ma,mi);
	}
	cout << ma << endl;
	return 0;
}