#include<bits/stdc++.h>
using namespace std;
const int N = 1e3+5;
const int INF = 1e9;
int p[N], x[N], y[N];
double distances[N], dis[N], maxdis[N];
double g[N][N];
char c[N][N];
int vis[N];
int n;

double f(int x1, int y1, int x2, int y2)
{
    return sqrt( 1.0 * (x1 - x2) * (x1 - x2) + 1.0 * (y1 - y2) * (y1 - y2)  );
}


void floyd()
{
    for(int k = 1; k <= n; k++){
        for(int i = 1; i <= n; i++){
            for(int j = 1; j <= n; j++){
                if(i != j && i != k && j != k && g[i][k] != INF && g[k][j] != INF)
                    if(g[i][j] > g[i][k] + g[k][j]) g[i][j] = g[i][k] + g[k][j];
            }
        }
    }
}

int main()
{
    scanf("%d", &n);
    for(int i = 1; i <= n; i++){
        scanf("%d%d", &x[i], &y[i]);
    }
    getchar();

    for(int i = 1; i <= n; i++){
        for(int j = 1; j <= n; j++){
            g[i][j] = INF;
        }
    }

    for(int i = 1; i <= n; i++){
        for(int j = 1; j <= n; j++){
            scanf("%c", &c[i][j]);
            if(c[i][j] == '1' ) {
                g[i][j] = f(x[i], y[i], x[j], y[j]);
            }
            if(i == j) g[i][j] = 0;
        }
        getchar();
    }

    floyd();


    ///找每个集合中，每个点到其他点的最短路的最大值
    ///maxdis[i]:以i为起点的最短路的最大值
    double ans1 = 0;
    for(int i = 1; i <= n; i++){
        maxdis[i] = 0;
        for(int j = 1; j <= n; j++){
            if(g[i][j] != INF && maxdis[i] < g[i][j])
            maxdis[i] = g[i][j];
        }
        ans1 = max(ans1, maxdis[i]);
    }

    ///找不在同一集合的点
    double ans2 = INF;
    for(int i = 1; i <= n; i++){
        for(int j = 1; j <= n; j++){
            if(g[i][j] == INF){
                if(f(x[i], y[i], x[j], y[j]) + maxdis[i] + maxdis[j] < ans2)
                    ans2 = f(x[i], y[i], x[j], y[j]) + maxdis[i] + maxdis[j];
            }
        }
    }
    if(ans1 < ans2) ans1 = ans2;
    printf("%.6f\n", ans1);
    return 0;
}