#include <bits/stdc++.h>
#define FOR(i,a,b) for(int i = a; i <= b; ++i)
#define ROF(i,a,b) for(int i = a; i >= b; --i)
using namespace std;

inline int rd() {
	int r = 0; bool w = false; char ch = getchar();
	while( ch < '0' || ch > '9' ) w = !(ch^45), ch = getchar();
	while( ch >= '0' && ch <= '9' ) r = (r<<1) + (r<<3) + (ch^48), ch = getchar();
	return w ? -r : r;
}

#define MAXN 10
const double eps = 1e-9;
int n;
int w[MAXN+5][MAXN+5];
double a[MAXN+5][MAXN+5];
void cal( int n ) { // n 个方程的高斯消元
    FOR(i,1,n) {
        int cons = i;
        FOR(j,i+1,n) {
            if( fabs(a[j][i]) > fabs(a[cons][i]) ) cons = j;
        }
        if( cons != i ) {
            for(int j = 1; j <= n+1; j++)
                swap(a[i][j], a[cons][j]);
        }
        if( fabs(a[i][i]) < eps ) {
            // puts("No Solution");
            return;
        }
        FOR(j,1,n) {
            if( j == i ) continue;
            double con = -a[j][i] / a[i][i];
            FOR(k,i,n+1) a[j][k] += con * a[i][k];
        }
    }
    FOR(i,1,n) {
        a[i][n+1] /= a[i][i];
    }
}
int popcount( int x ) {
    int ret = 0;
    while( x ) ret += x&1, x >>= 1;
    return ret;
}
int main() {
    n = rd();
    FOR(i,1,n) FOR(j,1,n) {
        w[i][j] = rd();
    }
    double ans = 0;
    int ag = 0;
    for(int s = 1; s < 1 << n; ++s) { // 枚举不为 0 的变量
        int prime = 0, eqcnt = 0;
        for(int i = 1; i <= n; ++i) if( (s >> (i-1)) & 1 ) {
            if( !prime ) {
                prime = i;
            } else {
                ++eqcnt;
                FOR(j,1,n+1) a[eqcnt][j] = 0;
                FOR(j, 1, prime-1) a[eqcnt][j] += w[j][prime];
                FOR(j, prime+1, n) a[eqcnt][j] += w[prime][j];
                FOR(j, 1, i-1) a[eqcnt][j] -= w[j][i];
                FOR(j, i+1, n) a[eqcnt][j] -= w[i][j];
            }
        } else {
            ++eqcnt;
            FOR(j, 1, n+1) a[eqcnt][j] = 0;
            a[eqcnt][i] = 1;
        }
        ++eqcnt;
        FOR(j, 1, n+1) a[eqcnt][j] = 1;
        cal(n);
        double now = 0; bool flag = true;
        FOR(i,1,n) {
            if( a[i][n+1] < -eps || a[i][n+1] > 1 + eps ) {
                flag = false;
                break;
            }
        }
        if( flag ) {
            FOR(i,1,n) FOR(j,i+1,n) now += a[i][n+1] * a[j][n+1] * w[i][j];
            if( now > ans ) ans = now, ag = s;
        }
    }
    // assert(ag + 1 == 1 << n);
    printf("%.7lf", ans);
    return 0;
}