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: Accepted)

#include<bits/stdc++.h>
#define For(i,a,b) for(i=(a);i<=(b);++i)
#define Forward(i,a,b) for(i=(a);i>=(b);--i)
#define Rep(i,a,b) for(register int i=(a);i<=(b);++i)
#define Repe(i,a,b) for(register int i=(a);i>=(b);--i)
using namespace std;
template<typename T>inline void read(T &x)
{
	T s=0,f=1;char k=getchar();
	while(!isdigit(k)&&(k^'-'))k=getchar();
	if(!isdigit(k)){f=-1;k=getchar();}
	while(isdigit(k)){s=s*10+(k^48);k=getchar();}
	x=s*f;
}
/*void init()
{
}
void solve()
{
}*/
const int MAXN=100;
int n,a[MAXN];
bool chk(int x)
{
	int last=a[1],in=0;
	Rep(i,2,n)
	{
		if(a[i]-last>=x)
		{
			last=a[i];
			continue;
		}
		if(a[i]-in<x)return false;
		in=a[i];
	}
	if(24-last-in<x)return false;
	return true;
}
int main()
{
	read(n);
	Rep(i,1,n)read(a[i]);
	sort(a+1,a+n+1);
	if(a[1]==0||n>=24)
	{
		printf("0\n");
		return 0;
	}
	if(n==1)
	{
		printf("%d\n",a[1]);
		return 0;
	}
	int l=1,r=a[1],mid=(l+r)>>1;
	while(l<=r)
	{
		if(chk(mid))l=mid+1;
		else r=mid-1;
		mid=(l+r)>>1;
	}
	printf("%d\n",r);
	//init();
	//solve();
	return 0;
}