1191

/*
DP
需要把棋盘割成n块，n>=2 && n<=14. 


稍加分析，可以得出，其实求的是平方和最小

写的时候确实比较繁琐。
*/


// include file
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <cctype>
#include <ctime>

#include <iostream>
#include <sstream>
#include <fstream>
#include <iomanip>
#include <bitset>

#include <algorithm>
#include <string>
#include <vector>
#include <queue>
#include <set>
#include <list>
#include <functional>

using namespace std;

// typedef
typedef long long LL;
typedef unsigned long long ULL;

// 
#define read freopen("in.txt","r",stdin)
#define write freopen("out.txt","w",stdout)
#define FORi(a,b,c) for(int i=(a);i<(b);i+=c)
#define FORj(a,b,c) for(int j=(a);j<(b);j+=c)
#define FORk(a,b,c) for(int k=(a);k<(b);k+=c)
#define FORp(a,b,c) for(int p=(a);p<(b);p+=c)
#define FORii(a,b,c) for(int ii=(a);ii<(b);ii+=c)
#define FORjj(a,b,c) for(int jj=(a);jj<(b);jj+=c)
#define FORkk(a,b,c) for(int kk=(a);kk<(b);kk+=c)

#define FF(i,a)    for(int i=0;i<(a);i++)
#define FFD(i,a)   for(int i=(a)-1;i>=0;i--)

#define Z(a) (a<<1)
#define Y(a) (a>>1)

const double eps = 1e-6;
const double INFf = 1e10;
const int INFi = 1000000000;
const double Pi = acos(-1.0);

template<class T> inline T sqr(T a){return a*a;}
template<class T> inline T TMAX(T x,T y)
{
	if(x>y) return x;
	return y;
}
template<class T> inline T TMIN(T x,T y)
{
	if(x<y) return x;
	return y;
}
template<class T> inline void SWAP(T &x,T &y)
{
	T t = x;
	x = y;
	y = t;
}
template<class T> inline T MMAX(T x,T y,T z)
{
	return TMAX(TMAX(x,y),z);
}
template<class T> inline T MMIN(T x,T y,T z)
{
	return TMIN(TMIN(x,y),z);
}


// code begin
int N,sum;
int mp[9][9];
int f[9][9][9][9]; //左上角，右下角
int DP[15][9][9][9][9]; //分割几次，左上角，右下角
int main()
{
	read;
	write;
	while(scanf("%d",&N)==1)
	{
		FORi(0,8,1)
		{
			FORj(0,8,1)
			{
				scanf("%d",&mp[i][j]);
				sum += mp[i][j];
			}
		}

		// 计算任意一个矩形的和
		int row;
		memset(f,0,sizeof(f));
		f[0][0][0][0] = mp[0][0];
		FORi(1,8,1)
		{
			f[0][0][0][i] = f[0][0][0][i-1]+mp[0][i];
		}
		FORi(1,8,1)
		{
			row = 0;
			FORj(0,8,1)
			{
				row+=mp[i][j];
				f[0][0][i][j] = f[0][0][i-1][j]+row;
			}
		}
		
		FORi(0,8,1)
		{
			FORj(0,8,1)
			{
				FORii(0,8,1)
				{
					FORjj(0,8,1)
					{
						if(i<=ii && j<=jj)
						{
							f[i][j][ii][jj] = f[0][0][ii][jj]-(j>=1?f[0][0][ii][j-1]:0)-(i>=1?f[0][0][i-1][jj]:0)+((i>=1&&j>=1)?f[0][0][i-1][j-1]:0);
						}
					}
				}
			}
		}
		FORi(0,8,1) FORj(0,8,1) FORii(0,8,1) FORjj(0,8,1) f[i][j][ii][jj]*=f[i][j][ii][jj];

		//开始DP
		int Min;

		FORi(0,8,1) FORj(0,8,1) FORii(0,8,1) FORjj(0,8,1) DP[0][i][j][ii][jj] = f[i][j][ii][jj];
		FORk(1,N+1,1)
		{
			FORi(0,8,1)
			FORj(0,8,1)
			FORii(0,8,1)
			FORjj(0,8,1)
			{
				if(i<=ii && j<=jj)
				{
					
					Min = 1000000000;
					//横划 
					FORp(i,ii,1)
					{
						Min = MMIN(Min,DP[k-1][i][j][p][jj]+f[p+1][j][ii][jj],DP[k-1][p+1][j][ii][jj]+f[i][j][p][jj]);
					}
					//
					FORp(j,jj,1)
					{
						Min = MMIN(Min,DP[k-1][i][j][ii][p]+f[i][p+1][ii][jj],DP[k-1][i][p+1][ii][jj]+f[i][j][ii][p]);
					}
					DP[k][i][j][ii][jj] = Min;
				}
			}
		}

		// DP[N][0][0][8][8]
		//printf("%f %f\n",DP[N-1][0][0][7][7]/(N+0.0),(sum*sum)/((N)*(N)+0.0));
		double ans = (DP[N-1][0][0][7][7]+0.0)/(N+0.0)-(sum*sum+0.0)/((N)*(N)+0.0) ;
		ans = sqrt(ans);
		printf("%.3f\n",ans);
	}
	return 0;
}