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 xx first
#define yy second
#define mp make_pair
#define pb push_back
#define mset(x, y) memset(x, y, sizeof x)
#define mcpy(x, y) memcpy(x, y, sizeof x)
using namespace std;

typedef long long LL;
typedef pair <int, int> pii;

inline int Read()
{
	int x = 0, f = 1, c = getchar();
	for (; !isdigit(c); c = getchar())
		if (c == '-')
			f = -1;
	for (;  isdigit(c); c = getchar())
		x = x * 10 + c - '0';
	return x * f;
}

const int MAXN = 55;

int n, ans, cnt[MAXN];
bool v[MAXN];

inline void DFS(int x)
{
	if (x == 13)
	{
		int ret = 24;
		for (int i = 0; i < 24; i ++)
			for (int j = i + 1; j < 24; j ++)
				if (v[i] && v[j])
					ret = min(ret, min(j - i, 24 - j + i));
		ans = max(ans, ret);
		return ;
	}
	if (cnt[x] == 2)
		v[x] = v[24 - x] = 1, DFS(x + 1);
	else if (cnt[x])
	{
		v[x] = 1, DFS(x + 1), v[x] = 0;
		if (x)
			v[24 - x] = 1, DFS(x + 1), v[24 - x] = 0;
	}
	else
		DFS(x + 1);
}

int main()
{
#ifdef wxh010910
	freopen("data.in", "r", stdin);
#endif
	n = Read();
	for (int i = 1; i <= n; i ++)
		cnt[Read()] ++;
	cnt[0] ++;
	for (int i = 1; i < 12; i ++)
		if (cnt[i] > 2)
			return puts("0"), 0;
	if (cnt[12] > 1 || cnt[0] > 1)
		return puts("0"), 0;
	DFS(0);
	return printf("%d\n", ans), 0;
}
