/*
dp[k][x1][y1][x2][y2]:左上角坐标为(x1,y1)，右下角坐标为(x2,y2)
的棋盘，设它把切割k次以后得到的k+1块矩形的总分平方和最小值.

s[x1][y1][x2][y2]:左上角坐标为(x1,y1)，右下角坐标为(x2,y2)
的棋盘的总和的平方

 dp[k][x1][y1][x2][y2] =
1)按横的划分： min(dp[k-1][x1][y1][f][y2]+s[f+1][y1][x2][y2]
    , dp[k-1][f+1][y1][x2][y2]+s[x1][y1][f][y2]);

2)按竖的划分： min(dp[k-1][x1][y1][x2][f]+s[x1][f+1][x2][y2]
    , dp[k-1][x1][f+1][x2][y2]+s[x1][y1][x2][f]);
*/

#include  < iostream >
#include  < cmath >
#define  INF 10000000
using   namespace  std;

const   int  m  =   8 ;
int  a[ 10 ][ 10 ];
int  n;
double  dp[ 16 ][ 10 ][ 10 ][ 10 ][ 10 ], s[ 10 ][ 10 ][ 10 ][ 10 ];

/**/ /*
dp[k][x1][y1][x2][y2]:左上角坐标为(x1,y1)，右下角坐标为(x2,y2)
的棋盘，设它把切割k次以后得到的k+1块矩形的总分平方和最小值.

s[x1][y1][x2][y2]:左上角坐标为(x1,y1)，右下角坐标为(x2,y2)
的棋盘的总和的平方


 dp[k][x1][y1][x2][y2] = 
1)按横的划分：    min(dp[k-1][x1][y1][f][y2]+s[f+1][y1][x2][y2]
                , dp[k-1][f+1][y1][x2][y2]+s[x1][y1][f][y2]);

2)按竖的划分：    min(dp[k-1][x1][y1][x2][f]+s[x1][f+1][x2][y2]
                , dp[k-1][x1][f+1][x2][y2]+s[x1][y1][x2][f]);
*/

double  mmin( double  a,  double  b)
{
     return  a  <  b  ?  a: b;
}

int  main()
{
     // freopen("data.txt", "r", std);

     int  x1, y1, x2, y2, t, f;
     double  sum, average, temp;
     while (cin  >>  n) {
        sum  =   0 ;
         for (x1  =   0 ; x1  <  m; x1 ++ )
             for (y1  =   0 ; y1  <  m; y1 ++ ) {
                cin  >>  a[x1][y1];
                sum  +=  a[x1][y1];
            }
        average  =  sum / n;

         for (x1  =   0 ; x1  <  m; x1 ++ ) {
             for (y1  =   0 ; y1  <  m; y1 ++ ) {
                 for (x2  =  x1; x2  <  m; x2 ++ ) {
                    sum  =   0 ;
                     for (y2  =  y1; y2  <  m; y2 ++ ) {
                        sum  +=  a[x2][y2];
                         if (x1  ==  x2)
                            s[x1][y1][x2][y2]  =  sum;
                         else
                            s[x1][y1][x2][y2]  =  s[x1][y1][x2 - 1 ][y2]  +  sum;
                    }
                }
            }
        }


         for (x1  =   0 ; x1  <  m; x1 ++ ) {
             for (y1  =   0 ; y1  <  m; y1 ++ ) {
                 for (x2  =  x1; x2  <  m; x2 ++ ) {
                     for (y2  =  y1; y2  <  m; y2 ++ ) {
                        s[x1][y1][x2][y2]  *=  s[x1][y1][x2][y2];
                        dp[ 0 ][x1][y1][x2][y2]  =  s[x1][y1][x2][y2];
                    }
                }
            }
        }

         for (t  =   1 ; t  <  n; t ++ ) {
             for (x1  =   0 ; x1  <  m; x1 ++ ) {
                 for (y1  =   0 ; y1  <  m; y1 ++ ) {
                     for (x2  =  x1; x2  <  m; x2 ++ ) {
                         for (y2  =  y1; y2  <  m; y2 ++ ) {
                            dp[t][x1][y1][x2][y2]  =  INF;
                             for (f  =  x1; f  <  x2; f ++ ) {
                                temp  =  mmin(dp[t - 1 ][x1][y1][f][y2] + s[f + 1 ][y1][x2][y2]
                                    , dp[t - 1 ][f + 1 ][y1][x2][y2] + s[x1][y1][f][y2]);
                                dp[t][x1][y1][x2][y2]  =  mmin(temp, dp[t][x1][y1][x2][y2]);
                            }
                             for (f  =  y1; f  <  y2; f ++ ) {
                                temp  =  mmin(dp[t - 1 ][x1][y1][x2][f] + s[x1][f + 1 ][x2][y2]
                                    , dp[t - 1 ][x1][f + 1 ][x2][y2] + s[x1][y1][x2][f]);
                                dp[t][x1][y1][x2][y2]  =  mmin(temp, dp[t][x1][y1][x2][y2]);
                            }
                        }
                    }
                }
            }
        }
        
         double  ans;
        ans  =  dp[n - 1 ][ 0 ][ 0 ][m - 1 ][m - 1 ] / n  -  average * average;
        printf( " %.3llf\n " , sqrt(ans));
    }
return   0 ;
}