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"
#define in std::cin
#define out std::cout
#define rep(i,N) for(int i=0;i<N;++i)
typedef long long int LL;

int N, ans;

int main()
{
	in >> N;
	std::vector<int>D(N);
	rep(i, N) in >> D[i];
	D.push_back(0); ++N;
	std::sort(D.begin(), D.end());
	rep(i, N)
	{
		if (i % 2) D[i] = 24 - D[i];
	}
	ans = 24;
	rep(i, N)
	{
		rep(j, N)
		{
			if (i == j) continue;
			auto d = abs(D[i] - D[j]);
			ans = std::min(ans, std::min(d, 24 - d));
		}
	}
	out << ans << std::endl;
	return 0;
}
