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 <stdio.h>
#include <vector>
#include <algorithm>
#include <string>
#include <cstring>
#include <cmath>
#include <map>
#include <queue>
#define maxn 50
int a[maxn];
using namespace std;

int main()
{
	int n,t;
	int MAX,MIN,t1,t2;
	while(~scanf("%d",&n)){
 
		for(int i=0;i<n;i++){
			scanf("%d",&a[i]);
	 
		}
		MAX=24;
		MAX=0;
		for(int i=0;i<n;i++){
			for(int j=0;j<2;j++){
				int MIN=24;
				for(int k=0;k<n;k++){
					if(k==i)continue;
					if(j==0){
						t1=fabs(a[j]-a[k]);
					}
					else 
						t1=fabs(24-a[j]-a[k]);
					MIN=min(t1,MIN);	
				}
				MAX=max(MAX,MIN);
			}
		}
// 		for(int j=0;j<n;j++)	
//		for(int k=j+1;k<n;k++){
//			t1=fabs(a[j]-a[k]);
//			t2=fabs(24-a[j]-a[k]);
//
//			MIN=min(MIN,max(t1,t2));
//	
//			
//		} 			
		if(n==1)
			MAX=0;

		printf("%d\n",MAX);
	}
}