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<bits/stdc++.h>
using namespace std;
int n,num[13],a[55],x[25],ans=0;
void check(){
	int w[25],d[13],tmp=0,Min=1e9;
	memset(w,0,sizeof (w));
	for(int i=1;i<=24;i++){
		//cout<<"X["<<i<<"]="<<x[i]<<"\n";
		if(x[i]==1){
			//cout<<"IN\n";
			w[++tmp]=i;
		}else if(x[i]>=2){
			return;
		}
	}
	/*cout<<"TIME:\n";
	for(int i=1;i<=tmp;i++){
		printf("%d ",w[i]);
	}
	puts("");*/
	w[0]=0;
	for(int i=1;i<=tmp;i++){
		Min=min(Min,w[i]-w[i-1]);
		//cout<<"MIN "<<Min<<"\n";
	}
	ans=max(ans,Min);
	//cout<<"ANS="<<ans<<"\n";
}
void dfs(int q){
	if(q==13){
	  check();
	  return ;
	}
	if(num[q]==1){
		x[q]++;
		//cout<<"INS "<<q<<"\n";
		dfs(q+1);
		x[q]--;
		//cout<<"ERA "<<q<<"\n";
		x[24-q]++;
		//cout<<"INS "<<24-q<<"\n";
		dfs(q+1);
		x[24-q]--;
		//cout<<"ERA "<<24-q<<"\n";
	}
	if(num[q]==2){
		x[q]++;
		//cout<<"INS "<<q<<"\n";
		x[24-q]++;
		//cout<<"INS "<<24-q<<"\n";
		dfs(q+1);
		x[q]--;
		//cout<<"ERA "<<q<<"\n";
		x[24-q]--;
		//cout<<"ERA "<<24-q<<"\n";
	}
	if(num[q]==0)
	  dfs(q+1);
}
int main(){
	memset(x,0,sizeof(x));
	scanf("%d",&n);
	for(int i=1;i<=n;i++){
	  scanf("%d",&a[i]);
	  num[a[i]]++;
	  if(num[a[i]]>2){
	  	puts("0");
	  	return 0;
	  }
	}
	if(num[0]){
		puts("0");
		return 0;
	}
	dfs(1);
	printf("%d",ans>=1e9?0:ans);
	return 0;
}