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 <cmath>
#include <string>
#include <vector>
#include <iomanip>
#include <iostream>
#include <algorithm>
#include <functional>
using namespace std;
int N, d, c[15], bit; bool f[29], ff[29];
int main() {
	cin >> N;
	for (int i = 0; i < N; i++) cin >> d, c[d]++;
	f[0] = true;
	bool flag = false;
	for (int i = 0; i <= 12; i++) {
		if (i == 0 && c[i] >= 1) flag = true;
		else if (i == 12 && c[i] >= 2) flag = true;
		else if (c[i] >= 3) flag = true;
		if (i == 12 && c[i] == 1) f[12] = true;
		if (1 <= i && i <= 11 && c[i] == 2) f[i] = f[24 - i] = true;
	}
	for (int i = 1; i < 12; i++) {
		if (c[i] == 1) bit |= 1 << i;
	}
	if (flag) cout << 0 << endl;
	else {
		int ret = 0;
		for (int i = 0; i < 1 << 12; i++) {
			if ((i | bit) != bit) continue;
			for (int j = 0; j < 24; j++) ff[j] = f[j];
			for (int j = 1; j < 12; j++) {
				if (c[j] != 1) continue;
				if (i & (1 << j)) ff[j] = true;
				else ff[24 - j] = true;
			}
			int res = 1 << 30;
			for (int j = 0; j < 24; j++) {
				for (int k = j + 1; k < 24; k++) {
					if (ff[j] && ff[k]) res = min(res, min(k - j, 24 - k + j));
				}
			}
			ret = max(ret, res);
		}
		cout << ret << endl;
	}
	return 0;
}