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
/*
  ID: oodt
  PROG:
  LANG:C++
*/
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cmath>
#include<string>
#include<cstring>
#include<map>
#include<vector>
#include<queue>
#include<stack>
#include<set>

using namespace std;

const int maxx=10005;
int n,m,k;
vector<int> a;
int ans = 0,cnt = 0,pos = 0;
int l = 0,r = 0;

int main()
{
#ifdef LOCAL
//    freopen("","r",stdin);
#endif
    scanf("%d",&n);
    for(int i = 0; i < n; i++)
    {
        int x;
        scanf("%d",&x);
        a.push_back(x);
    }
    a.push_back(0);
    int mi = 24;
    sort(a.begin(),a.end());
    for(int i = 1; i <= n; i++)
    {
        if(i%2) a[i] *= -1;
    }
    sort(a.begin(),a.end());
    for(int i = 1; i <= n; i++)
    {
        mi = min(mi,a[i]-a[i-1]);
    }
    mi = min(mi,a[0]+24-a[n]);
    printf("%d\n",mi);
    return 0;
}
