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>
using namespace std;
#include<string.h>
#include<set>
#include<stdio.h>
#include<math.h>
#include<queue>
#include<map>
#include<algorithm>
#include<cstdio>
#include<cmath>
#include<cstring>
#include <cstdio>
#include <cstdlib>
#include<stack>
#include<vector>
int  a[100];
int b[100];
int main()
{
    int n;
    cin>>n;
    for(int i=1;i<=n;i++)
        cin>>a[i];
    if(n==1)
    {
        cout<<a[1]<<endl;
        return 0;
    }
    int min1=1e9;
    int min2=1e9;
    b[0]=0;
    sort(a+1,a+1+n);
    for(int i=1;i<=n;i++)
    {
        if(i%2==0)
            b[i]=24-a[i];
        else
            b[i]=a[i];
    //cout<<b[i]<<endl;
    }
    sort(b,b+1+n);
    for(int i=1;i<=n;i++)
    {
         min1=min(min1,abs(b[i]-b[i-1]));
         //cout<<b[i]<<endl;
    }
     for(int i=1;i<=n;i++)
    {
        if(i%2==1)
            b[i]=24-a[i];
        else
            b[i]=a[i];
            //cout<<b[i]<<endl;
    }
    sort(b,b+1+n);
    for(int i=1;i<=n;i++)
    {
        min2=min(min2,abs(b[i]-b[i-1]));
    }
    cout<<min(min1,min2)<<endl;
}
