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 <map>
#include <list>
#include <cmath>
#include <cstdio>
#include <vector>
#include <cstring>
#include <iostream>
#include <algorithm>
#define LL long long
using namespace std;

int cal[13] = {0};
int val[26] = {0},vc;
int a[55] = {0};
int n,tmp,maxn = 0;

int main(){
	scanf("%d",&n);
	for(register int i = 0;i < n;i ++){
		scanf("%d",&a[i]);
		cal[a[i]] ++;
		if(cal[a[i]] >= 3 || cal[12] >= 2 || cal[0] >= 1){
			printf("0\n");
			return 0;
		}
	}
	for(register int i = 0;i < (1 << 22);i ++){
		memset(val,0,sizeof(val)); vc = 0;
		for(register int j = 1;j <= 11;j ++){
			if(cal[j] == 0) continue;
			if(cal[j] == 1){
				if((i >> (j - 1 + j - 1)) & 1) val[++ vc] = j;
				else val[++ vc] = 24 - j;
			}
			if(cal[j] == 2){
				if((i >> (j - 1 + j - 1)) & 1) val[++ vc] = j;
				else val[++ vc] = 24 - j;
				if((i >> (j - 1 + j)) & 1) val[++ vc] = j;
				else val[++ vc] = 24 - j;
			}
		}
		if(cal[12]) val[++ vc] = 12;
		val[++ vc] = 24;
		sort(val + 1,val + 1 + vc);
		tmp = 24;
		for(register int j = vc;j >= 1;j --) tmp = min(tmp,val[j] - val[j - 1]);
		maxn = max(maxn,tmp);
	}
	printf("%d\n",maxn);
	return 0;
}