題意：
　　給一個圖，每條邊有出現概率，求這個圖恰好為一棵樹的概率。
　
解法：
　　考慮Kirchhoff矩陣的意義：K[G]=D[G]−A[G]=B[G]B[G]T，之所以能夠進行生成樹計數是對於其伴隨矩陣在計數n−1條邊的集合時，當n−1條邊中存在環就會產生線性組合而導致行列式為零，否則恰好對角線上均為伴隨矩陣中所賦的值，使得det(Bi,j)2就為1。
　　因此我們可以令鄰接矩陣中存下邊權，使得可以用Kirchhoff矩陣計數生成樹中邊權的乘積。
　　考慮到答案為∑T是一棵生成樹∏u,v(p(u,v)∗[(u,v)∈T])((1−p(u,v))∗[u,v∉T])，我們可以令邊權為p

/*
    I will chase the meteor for you, a thousand times over.
    Please wait for me, until I fade forever.
    Just 'coz GEOTCBRL.
*/
 

#include <bits/stdc++.h>
using namespace std;
#define fore(i,u)  for (int i = head[u] ; i ; i = nxt[i])
#define rep(i,a,b) for (int i = a , _ = b ; i <= _ ; i ++)
#define per(i,a,b) for (int i = a , _ = b ; i >= _ ; i --)
#define For(i,a,b) for (int i = a , _ = b ; i <  _ ; i ++)
#define Dwn(i,a,b) for (int i = ((int) a) - 1 , _ = (b) ; i >= _ ; i --)
 

#define fors(s0,s) for (int s0 = (s) , _S = s ; s0 ; s0 = (s0 - 1) & _S)
#define foreach(it,s) for (__typeof(s.begin()) it = s.begin(); it != s.end(); it ++)

#define mp make_pair
#define pb push_back
#define pii pair<int , int>
#define fir first
#define sec second
#define MS(x,a) memset(x , (a) , sizeof (x))

#define gprintf(...) fprintf(stderr , __VA_ARGS__)
#define gout(x) std::cerr << #x << "=" << x << std::endl
#define gout1(a,i) std::cerr << #a << '[' << i << "]=" << a[i] << std::endl
#define gout2(a,i,j) std::cerr << #a << '[' << i << "][" << j << "]=" << a[i][j] << std::endl
#define garr(a,l,r,tp) rep (__it , l , r) gprintf(tp"%c" , a[__it] , " \n"[__it == _])

template <class T> inline void upmax(T&a , T b) { if (a < b) a = b ; }
template <class T> inline void upmin(T&a , T b) { if (a > b) a = b ; }

typedef long long ll;

const int maxn = 57;
const int maxm = 200007;
const int mod = 1000000007;
const int inf = 0x7fffffff;
const double eps = pow(2 , -19);

typedef int arr[maxn];
typedef int adj[maxm];

inline int fcmp(double a , double b) {
    if (fabs(a - b) <= eps) return 0;
    if (a < b - eps) return -1;
    return 1;
}

inline int add(int a , int b) { return ((ll) a + b) % mod ; }
inline int mul(int a , int b) { return ((ll) a * b) % mod ; }
inline int dec(int a , int b) { return add(a , mod - b % mod) ; }
inline int Pow(int a , int b) {
    int t = 1;
    while (b) {
        if (b & 1) t = mul(t , a);
        a = mul(a , a) , b >>= 1;
    }
    return t;
}

#define rd RD<int>
#define rdll RD<long long>
template <typename Type>
inline Type RD() {
    Type x = 0;
    int flag = 0;
    char c = getchar();
    while (!isdigit(c) && c != '-')
        c = getchar();
    (c == '-') ? (flag = 1) : (x = c - '0');
    while (isdigit(c = getchar()))
        x = x * 10 + c - '0';
    return flag ? -x : x;
}
inline char rdch() {
    char c = getchar();
    while (!isalpha(c)) c = getchar();
    return c;
}

// beginning

int n;
double to_mul , a[maxn][maxn];

void input() {
    n = rd();
    to_mul = 1;
    rep (i , 1 , n) rep (j , 1 , n) {
        scanf("%lf" , &a[i][j]);
        double b = 1 - a[i][j];
        upmax(b , eps);
        if (i < j)
            to_mul *= b;
        if (i != j)
            a[i][j] = - a[i][j];
        a[i][j] /= b;
    }
    rep (i , 1 , n) rep (j , i + 1 , n)
        a[i][i] -= a[i][j] , a[j][j] -= a[i][j];
}

inline double det() {
    double ret = 1;
    For (i , 1 , n) {
        int j = i;
        while (j < n && fabs(a[j][i]) < eps) j ++;
        if (j == n) return 0;
        if (i != j) {
            ret *= -1;
            For (k , 1 , n) swap(a[i][k] , a[j][k]);
        }
        For (j , i + 1 , n) if (fabs(a[j][i]) >= eps) {
            double t = a[j][i] / a[i][i];
            For (k , i , n) a[j][k] -= t * a[i][k];
        }
    }
    For (i , 1 , n)
        ret *= a[i][i];
    return fabs(ret);
}

void solve() {
    double ans = det();
    ans *= to_mul;
    printf("%.10lf\n" , ans);
}

int main() {
    #ifndef ONLINE_JUDGE
        freopen("data.txt" , "r" , stdin);
    #endif
    input();
    solve();
    return 0;
}