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 <bits/stdc++.h>
using namespace std;
const int INF=0x3f3f3f3f;
//#define mod 1000000007
#define PI acos(-1.0)
//const long long MAX=200005;
typedef long long ll;
//int d[8][2]={-1,0,0,-1,0,1,1,0,1,1,-1,-1,-1,1,1,-1};
int n,m,t=0;
char s[100005];
int a[100],b[100];
int MIN=24,MAX=24;
int main()
{
    a[0]=0;
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
        scanf("%d",&a[i]);
    if(n==1)
    {
        printf("%d\n",a[1]);
        return 0;
    }
    sort(a+1,a+n+1);
    for(int i=n;i>0;i--)
    {
        if(i%2==0)
        b[i]=24-a[i];
        else b[i]=a[i];
    }
    b[0]=0;
    sort(b,b+n+1);
    for(int i=1;i<=n;i++)
    {
        if(b[i]-b[i-1]<MIN)
            MIN=b[i]-b[i-1];
    }
    printf("%d\n",MIN);
    return 0;
}
