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<string>
#include<algorithm>
#include <string.h>
#include <stdio.h>
#include <math.h>
#include <set>
#include <queue>
#include <stack>
#include <map>

using namespace std;
typedef int status;
typedef long long LL;
const int INF = int(1e9);
const int MAX = 400002;

int N;
int a[MAX];

int main()
{
    cin>>N;
    for(int i = 1;i<=N;i++)
        cin>>a[i];
    sort(a+1,a+1+N);

    for(int i = 1;i<=N;i++)
    {
        if(i%2==0)
        {
            a[i] = 24 - a[i];
        }
    }
    sort(a+1,a+1+N);

    int maxn = 25;
    for(int i = 1;i<=N;i++)
    {
        maxn = min(maxn,a[i] - a[i-1]);
    }

    cout<<maxn<<endl;

    return 0;
}
