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 ;
int a[55] ;
int b[15] ;
int q[100] , r[100] ;
int ans , cnt ;
void dfs( int k ){
     if ( k == 13 ){
         for( int i = 1 ; i <= cnt ; i++ )
             r[i] = q[i] ;
         sort( r + 1 , r + 1 + cnt ) ;
         int minx = 10086 ;
         for( int i = 1 ; i <= cnt - 1 ; i++ )
             minx = min( minx , r[i + 1] - r[i] ) ;
         minx = min( minx , 24 - r[cnt] + r[1] ) ;
         ans = max ( ans , minx ) ;
         return ;
     }
     if ( b[k] == 0 ) dfs( k + 1 );
     if ( b[k] == 1 ){
         cnt++ ; q[cnt] = k ; dfs( k + 1 ) ; cnt-- ;
         cnt++ ; q[cnt] = 24 - k ; dfs( k + 1 ) ; cnt-- ;
     }
     if ( b[k] == 2 ){
         cnt++ ; q[cnt] = k ; cnt++ ;
         q[cnt] = k ; dfs( k + 1 ) ; cnt -= 2 ;
         cnt++ ; q[cnt] = 24 - k ; cnt++ ;
         q[cnt] = k ; dfs( k + 1 ) ; cnt -= 2 ;
         cnt++ ; q[cnt] = 24 - k ; cnt++ ;
         q[cnt] = 24 - k ; dfs( k + 1 ) ; cnt -= 2 ;
     }
}
int main(){
    int N ;
    cin >> N ;
    for( int i = 1 ; i <= N ; i++ ){
        cin >> a[i] ;
        b[a[i]]++ ;
    }
    for( int i = 0 ; i <= 12 ; i++ ){
        if ( b[i] > 2 ){
            cout << 0 << endl ;
            return 0 ;
        }
    }
    cnt = 1 ; q[1] = 0 ;
    if ( b[0] ){
        cout << 0 << endl ;
        return 0 ;
    }
    dfs( 1 ) ;
    cout << ans << endl ;
    return 0 ;
}