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 <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <vector>
#define repi(i,a,b) for(int i=(a);i<(b);++i)
#define rep(i,a) repi(i,0,a)
#define clr(a,v) memset((a),(v),sizeof(a))

int N, D[55], b[55];
int cnt[55];

int main()
{
  scanf( "%d", &N );
  rep( i, N )
    scanf( "%d", D+i ), ++cnt[D[i]];

  int ans = 1;

  if( N >= 14 )
  {
    if( cnt[0] >= 1 || cnt[12] >= 2 )
      ans = 0;

    repi( i, 1, 12 ) if( cnt[i] >= 3 )
      ans = 0;
  }
  else
  {
    ans = 0;
    D[N] = 0;
    for( int i = 0; i <= (1<<14); ++i )
    {
      int mi = 1<<30;
      rep( j, N+1 )
        b[j] = D[j]+(i>>j&1?24-2*D[j]:0);

      rep( j, N+1 ) rep( k, N+1 ) if( j != k )
        mi = std::min( mi, std::min( abs(b[k]-b[j]), 24-abs(b[k]-b[j]) ) );

      ans = std::max( ans, mi );
    }
  }

  if( N == 1 )
    ans = std::min( D[0], 24-D[0] );

  printf( "%d\n", ans );

  return 0;
}