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<iostream>
#include<algorithm>
#include<vector>
#define LL long long 
int N;
int T[13];
bool used[48];
int ans;
std::vector<int> t;
void dfs(int n) {
	if (n == t.size()) {
		int res = 100;
		std::vector<int> copy = t;
		std::sort(copy.begin(), copy.end());
		for (int i = 1; i < copy.size(); i++) {
			res = std::min(res, copy[i] - copy[i - 1]);
		}
		res = std::min(res, copy[0] + 24 - copy[copy.size()-1]);
		ans = std::max(res, ans);
		return;
	}
	if (T[t[n]] == 2) {
		dfs(n + 2);
		return;
	}
	if (t[n] == 12) {
		dfs(n + 1);
		return;
	}
	dfs(n + 1);
	t[n] = 24 - t[n];
	dfs(n + 1);
	t[n] = 24 - t[n];
	return;
}
int main() {
	T[0]++;
	std::cin >> N;
	if (N == 1) {
		int tmp;
		std::cin >> tmp;
		std::cout << tmp << std::endl;
		return 0;
	}
	for (int i = 1; i <= N; i++) {
		int tmp;
		std::cin >> tmp;
		T[tmp]++;
	}
	if (T[0] > 1 || T[12] > 1) {
		std::cout << 0 << std::endl;
		return 0;
	}
	for (int i = 0; i <= 12; i++) {
		if (T[i] >= 1)t.emplace_back(i);
		if (T[i] >= 2)t.emplace_back(24 - i);
		if (T[i] >= 3) {
			std::cout << 0 << std::endl;
			return 0;
		}
	}
	dfs(1);
	std::cout << ans << std::endl;
	return 0;
}