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<cstdio>
#include<cmath>
#include<algorithm>
#define N 100
using namespace std;
int a[N];
int n;
int main()
{
    cin>>n;
    for (int i=1;i<=n;i++)
        cin>>a[i];
    sort(a+1,a+1+n);
    for (int i=n;i>=1;i--)
        if (i%2==0)
          a[i]=24-a[i];
    sort(a+1,a+1+n);
    int maxn=24;
    for (int i=1;i<=n;i++)
        maxn=min(maxn,a[i]-a[i-1]);
    cout << maxn << endl;
    return 0;
}
