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.

Examples

Input

3
7 12 8


Output

4


Input

2
11 11


Output

2


Input

1
0


Output

0
#include<cstdio>
#include<algorithm>
#include<cmath>
using namespace std;

int f[24],c[13];

int dfs(int x){
    int ans=24;
    if(x==12){
        for(int i=0;i<24;i++){
            for(int j=i+1;j<24;j++){
                if(f[i]&&f[j]){
                    ans=min(ans,min(j-i,24-j+i));
                }
            }
        }
        return ans;
    }
    if(c[x]!=1){
        return dfs(x+1);
    }
    f[x]=1;
    ans=dfs(x+1);
    f[x]=0;
    f[24-x]=1;
    ans=max(ans,dfs(x+1));
    f[24-x]=0;
    return ans;
}

int main(){
    int n,d;
    f[0]=1;
    scanf("%d",&n);
    for(int i=0;i<n;i++){
        scanf("%d",&d);
        c[d]++;
    }
    if(c[0]){
        printf("0\n");
        return 0;
    }
    for(int i=1;i<13;i++){
        if(c[i]==2){
            if(i==12){
                printf("0\n");
                return 0;
            }
            f[i]=1;
            f[24-i]=1;
        }
        if(c[i]>2){
            printf("0\n");
            return 0;
        }
    }
    f[12]=c[12];
    printf("%d\n",dfs(1));
}