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<iostream>
#include<algorithm>
#include<vector>
#include<string>
#include<utility>
#include<map>
#include<set>
#include<queue>
#include<math.h>
using namespace std;
#define mod (1000000000+7)
#define N (10007)
#define INF 1e16
typedef long long ll;
typedef pair<ll,ll> P;

ll cnt[13];
ll used[24];

int main(void){
    int n,d;
    int ans=0,num=24;
    cin>>n;
    cnt[0]++;
    for(int i=0;i<n;i++){
        cin>>d;
        cnt[d]++;
    }
    for(int i=0;i<(1<<11);i++){
        num=24;
        for(int j=0;j<24;j++)used[j]=false;
        for(int j=0;j<13;j++){
            if(cnt[j]==0)continue;
            if(j==0 || j==12){
                used[j]=true;
                if(cnt[j]>=2)num=0;
            }
            else{
                if(cnt[j]>=3)num=0;
                else{
                    if(cnt[j]==2){
                        used[j]=true;
                        used[24-j]=true;
                    }
                    else{
                        used[i&(1<<(j-1))?j:24-j]=true;
                    }
                }
            }
        }
        for(int i=0;i<23;i++){
            for(int j=0;j<23;j++){
                if(i==j)continue;
                if(used[i]&used[j]){
                    int x=i-j;
                    if(x<0)x*=-1;
                    num=min(num,min(x,24-x));
                }
            }
        }
        ans=max(ans,num);
    }
    cout<<ans<<endl;
    return 0;
}