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;

void solve()
{
	int N;
	cin >> N;
	vector<int> s(12);
	int ans = 0;

	for (int i = 0; i < N; i++)
	{
		int d;
		cin >> d;
		if (d)
		{
			s[d-1]++;
			if (s[d-1]==3)
			{
				cout << 0 << endl;
				return;
			}
		}
		else
		{
			cout << 0 << endl;
			return;
		}
	}
	for (int bit = 0; bit < 1<<12; bit++)
	{
		vector<int> v;
		v.push_back(0);
		for (int i = 0; i < 12; i++)
		{
			if (!s[i]) continue;
			v.push_back((bit>>i&1)?i+1:23-i);
			if (s[i]>1) v.push_back((bit>>i&1)?23-i:i+1);
		}
		int n = v.size(), res = 12;
		for (int i = 0; i < n-1; i++)
			for (int j = i+1; j < n; j++)
			{
				int diff = abs(v[j]-v[i]);
				res = min(res, min(24-diff, diff));
			}
		ans = max(ans, res);
	}
	cout << ans << endl;
}

int main(void)
{
	solve();
	//cout << "yui(*-v・)yui" << endl;
	return 0;
}
