#include <iostream>
#include <cmath>

using namespace std;

const int N = 9, M = 15;
const double INF = 1e8;

int n;
double f[N][N][N][N][M];
int s[N][N]; //前缀和数组

//获取局部sqrt(x) / n
double get(int x1, int y1, int x2, int y2) {
    double sum = s[x2][y2] - s[x1 - 1][y2] - s[x2][y1 - 1] + s[x1 - 1][y1 - 1];
    return sum * sum / n;
}

double dp(int x1, int y1, int x2, int y2, int k) {
    double &t = f[x1][y1][x2][y2][k];

    if (t) return t; //记忆化搜索

    if (k == 1) //不需要再分割
    {
        t = get(x1, y1, x2, y2);
        return t;
    }

    //枚举所有分割方案
    t = INF;
    for (int x = x1; x < x2; x++) //横向切割
    {
        t = min(t, dp(x1, y1, x, y2, 1) + dp(x + 1, y1, x2, y2, k - 1));
        t = min(t, dp(x1, y1, x, y2, k - 1) + dp(x + 1, y1, x2, y2, 1));
    }

    for (int y = y1; y < y2; y++) //纵向切割
    {
        t = min(t, dp(x1, y1, x2, y, 1) + dp(x1, y + 1, x2, y2, k - 1));
        t = min(t, dp(x1, y1, x2, y, k - 1) + dp(x1, y + 1, x2, y2, 1));
    }

    return t;
}

int main() {
    cin >> n;

    for (int x = 1; x <= 8; x++)
        for (int y = 1; y <= 8; y++) {
            cin >> s[x][y];
            s[x][y] += s[x - 1][y] + s[x][y - 1] - s[x - 1][y - 1];
        }
    // 这段注释掉
    // for (int x1 = 1; x1 <= 8; x1++)
    //     for (int y1 = 1; y1 <= 8; y1++)
    //         for (int x2 = 1; x2 <= 8; x2++)
    //             for (int y2 = 1; y2 <= 8; y2++)
    //                 for (int k = 1; k <= n; k++)
    //                     f[x1][y1][x2][y2][k] = INF;

    printf("%.3lf\n", sqrt(dp(1, 1, 8, 8, n) - get(1, 1, 8, 8) / n));

    return 0;
}