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 <cstdio>
#include <algorithm>
#include <queue>
#include <cmath>
#define ROP(i,n) for(int i=0;i<n;i++)
using namespace std;
priority_queue<int> que;
int main(){
  int N,D[50],DD[50],cnt=0;
  scanf("%d",&N);
  ROP(i,N){
    scanf(" %d",&D[i]);
    DD[i]=D[i];
  }
  if(N==1){
    printf("0\n");return 0;
  }
  ROP(i,N){
    if(D[i]==12) cnt++;
    else if(cnt==2){
      printf("0\n");return 0;
    }else if(D[i]==0){
      printf("0\n");return 0;
    }
  }
  ROP(i,N){
    int MIN=12,cnt=0;DD[i]=24-D[i];
    ROP(i,N){
      MIN=min(MIN,D[i]);
    }
    ROP(j,N){
      if(i==j) continue;
      MIN=min(MIN,abs(DD[i]-D[j]));
    }
    DD[i]=D[i];
    ROP(j,N){
      if(i==j) continue;
      if(DD[i]==D[j]){
        if(++cnt==2){
          printf("0\n");
          return 0;
        }
        continue;
      }
      MIN=min(MIN,abs(DD[i]-D[j]));
    }
    que.push(MIN);
  }
  printf("%d\n",que.top());
  return 0;
}