Problem:
Score : 500 points

Problem Statement

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.

Sample Input 1

3
7 12 8

Sample Output 1

4

For example, consider the situation where it is 7, 12 and 16 o'clock in each person's city at the moment when it is 0 o'clock in Takahashi's city. In this case, the time gap between the second and third persons' cities is 4 hours.

Sample Input 2

2
11 11

Sample Output 2

2

Sample Input 3

1
0

Sample Output 3

0

Note that Takahashi himself is also a participant.

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<iostream>
#include<cmath>
#include<algorithm>
using namespace std;
int cnt[24]={0};
bool flag=false;
bool v[25];
int res=-1;
void find(int x)
{
	if(x==13)
	{
		int a=24;
		for(int i=0;i<24;i++)
			for(int j=i+1;j<24;j++)
				if(v[i]&&v[j])
					a=min(a,min(j-i,24-j+i));
		res=max(a,res);
		return;
	}
	if(cnt[x]==0) find(x+1);
	else if(cnt[x]==2) 
	{
		v[x]=v[24-x]=1;
		find(x+1);
	}
	else
	{
		v[x]=1;
		find(x+1);
		v[x]=0;
		if(x)
		{
			v[24-x]=1;
			find(x+1);
			v[24-x]=0;
		}
	}
}
int main()
{
	int n,m;
	scanf("%d",&n);
	cnt[0]=1;
	for(int i=1;i<=n;i++)
	{
		scanf("%d",&m);
		cnt[m]++;
		if(m) cnt[24-m]++;
		if(cnt[m]>2) flag=true;
	}
	if(flag)
	{
		cout<<0<<endl;
		return 0;
	}
	else
	{
		find(1);
		cout<<res<<endl;
	}
	return 0;
}