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<fstream>
#include<algorithm>
#include<string>
#include<cstring>

using namespace std;

const int MAXN = 50 + 5;

int N;
int Ans = 0;
int A[MAXN], B[MAXN], Used[MAXN];



void DFS(int depth) {
	if (depth == N + 1) {
		int C[MAXN];
		memcpy(C, B, sizeof(int) * (N + 1));
		sort(C, C + N + 1);
		int tmp = C[1] - C[0];
		for (int i = 1; i < N + 1; i++) {
			if (tmp > C[i] - C[i - 1]) {
				tmp = C[i] - C[i - 1];
			}
		}
		tmp = min(tmp, 24 - C[N]);
		if (Ans < tmp)
			Ans = tmp;
		return;
	}
	if (Used[depth]) {
		DFS(depth + 1);
		return;
	}
		

	B[depth] = A[depth];
	DFS(depth + 1);
	B[depth] = 24 - A[depth];
	DFS(depth + 1);
	return;
}

int main() {
	cin >> N;
	for (int i = 0; i < N; i++)
		cin >> A[i];

	bool flag = true;
	if (N >= 24) cout << 0 << endl;
	else {
		A[N] = 0;
		sort(A, A + N + 1);
		if (A[1] == 0)
			cout << 0 << endl;
		else {
			for (int i = 0; i < N; i++) {
				if (A[i] == A[i + 1]) {
					if (Used[i]) {
						cout << 0 << endl;
						flag = false;
						break;
					}
					Used[i] = 1;
					Used[i + 1] = 1;
					B[i] = A[i];
					B[i + 1] = 24 - A[i];
				}
			}
			if (flag) {
				DFS(1);
				cout << Ans << endl;
			}
			
		}
	}
	return 0;
}