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 <iostream>
#include <algorithm>
#include <cstring>
using namespace std;
int n;
int d[100];
int t[13];
int mo[25];
int dfs(int a){
    if(a==n){
        int pre=0,ans=24;
        for(int i=1;i<=24;i++){
            if(mo[i%24]){
                ans=min(i-pre,ans);
                pre=i;
            }
        }
        return ans;
    }
    int x=0,y=0;
    if(mo[d[a]]==0){
        mo[d[a]]=1;
        x=dfs(a+1);
        mo[d[a]]=0;
    }
    if(mo[24-d[a]]==0){
        mo[24-d[a]]=1;
        y=dfs(a+1);
        mo[24-d[a]]=0;
    }
    return max(x,y);
}
int main(void){
    cin>>n;
    t[0]=2;
    for(int i=0;i<n;i++){
        cin>>d[i];
        if(++t[d[i]]>2){
            cout<<0<<endl;
            return 0;
        }
    }
    mo[0]=1;
    cout<<dfs(0)<<endl;
}
