#include<bits/stdc++.h>
#define all(x) x.begin(), x.end()
#define ff first
#define ss second
#define O_O
using namespace std;
template <typename T>
using bstring = basic_string<T>;
template <typename T>
using matrix = vector<vector<T>>;
typedef unsigned int uint;
typedef unsigned long long ull;
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
typedef double dbl;
typedef long double dbll; 
const ll INFL = 4e18+25;
const int INF = 1e9+42;
const double EPS = 1e-7;
const int MOD = (1<<23)*17*7 + 1; // 998244353
const int RANDOM = chrono::high_resolution_clock::now().time_since_epoch().count();
const int MAXN = 1e6+1;

/**
 * Author: Stanford
 * Source: Stanford Notebook
 * License: MIT
 * Description: Solves a general linear maximization problem: maximize $c^T x$ subject to $Ax \le b$, $x \ge 0$.
 * Returns -inf if there is no solution, inf if there are arbitrarily good solutions, or the maximum value of $c^T x$ otherwise.
 * The input vector is set to an optimal $x$ (or in the unbounded case, an arbitrary solution fulfilling the constraints).
 * Numerical stability is not guaranteed. For better performance, define variables such that $x = 0$ is viable.
 * Usage:
 * vvd A = {{1,-1}, {-1,1}, {-1,-2}};
 * vd b = {1,1,-4}, c = {-1,-1}, x;
 * T val = LPSolver(A, b, c).solve(x);
 * Time: O(NM * \#pivots), where a pivot may be e.g. an edge relaxation. O(2^n) in the general case.
 * Status: seems to work?
 */
#pragma once
#define rep(_i,_j,_k) for(int _i = _j; _i < _k; _i++)

typedef long double T; // long double, Rational, double + mod<P>...
typedef vector<T> vd;
typedef vector<vd> vvd;

const T eps = 1e-8, inf = 1/.0;
#define MP make_pair
#define ltj(X) if(s == -1 || MP(X[j],N[j]) < MP(X[s],N[s])) s=j

using vi = vector<int>;
#define sz(x) (int)x.size()

struct LPSolver {
	int m, n;
	vi N, B;
	vvd D;

	LPSolver(const vvd& A, const vd& b, const vd& c) :
		m(sz(b)), n(sz(c)), N(n+1), B(m), D(m+2, vd(n+2)) {
            // for(int i = 0; i < A.size(); i++){
                // for(int j = 0; j < A[0].size(); j++){
                //     cerr << A[i][j] << ' ';
                // }
                // cerr << " <= " << b[i];
                // cerr << '\n';
            // }


			rep(i,0,m) rep(j,0,n) D[i][j] = A[i][j];
			rep(i,0,m) { B[i] = n+i; D[i][n] = -1; D[i][n+1] = b[i];}
			rep(j,0,n) { N[j] = j; D[m][j] = -c[j]; }
			N[n] = -1; D[m+1][n] = 1;
		}

	void pivot(int r, int s) {
		T *a = D[r].data(), inv = 1 / a[s];
		rep(i,0,m+2) if (i != r && abs(D[i][s]) > eps) {
			T *b = D[i].data(), inv2 = b[s] * inv;
			rep(j,0,n+2) b[j] -= a[j] * inv2;
			b[s] = a[s] * inv2;
		}
		rep(j,0,n+2) if (j != s) D[r][j] *= inv;
		rep(i,0,m+2) if (i != r) D[i][s] *= -inv;
		D[r][s] = inv;
		swap(B[r], N[s]);
	}

	bool simplex(int phase) {
		int x = m + phase - 1;
		for (;;) {
			int s = -1;
			rep(j,0,n+1) if (N[j] != -phase) ltj(D[x]);
			if (D[x][s] >= -eps) return true;
			int r = -1;
			rep(i,0,m) {
				if (D[i][s] <= eps) continue;
				if (r == -1 || MP(D[i][n+1] / D[i][s], B[i])
				             < MP(D[r][n+1] / D[r][s], B[r])) r = i;
			}
			if (r == -1) return false;
			pivot(r, s);
		}
	}

	T solve(vd &x) {
		int r = 0;
		rep(i,1,m) if (D[i][n+1] < D[r][n+1]) r = i;
		if (D[r][n+1] < -eps) {
			pivot(r, n);
			if (!simplex(2) || D[m+1][n+1] < -eps) return -inf;
			rep(i,0,m) if (B[i] == -1) {
				int s = 0;
				rep(j,1,n+1) ltj(D[i]);
				pivot(i, s);
			}
		}
		bool ok = simplex(1); x = vd(n);
		rep(i,0,m) if (B[i] < n) x[B[i]] = D[i][n+1];
		return ok ? D[m][n+1] : inf;
	}
};

int main(){
    
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    
    int n;
    cin >> n;

    vvd A(2*n,vd(n));

    matrix<dbll> v(n,vector<dbll>(n));
    for(int i = 0; i < n; i++){
        for(int j = 0; j < n; j++)
            cin >> v[i][j];
    }

    for(int i = 0; i < n; i++){
        for(int j = 0; j < n; j++){
            A[2*i][j] = v[i][j]-v[(i+1)%n][j];
            A[2*i+1][j] = -(v[i][j]-v[(i+1)%n][j]);
        }
    }
    
    A.push_back(vd(n,1));
    A.push_back(vd(n,-1));

    vd B(A.size());

    B.back() = -1;
    B[A.size()-2] = 1;

    vd C = v[0];

    vd resp(n);
    resp[0] = 1;
    cout << fixed << setprecision(7) << LPSolver(A,B,C).solve(resp)/2 << '\n';
    
    // for(int i = 0; i < n; i++)
    //     cerr << resp[i] << ' ';
    return 0;

}