#include <bits/stdc++.h>
using namespace std;
const int MAXN = 150;
const double INF = 200000 * 150;
int n, m, s, t;
int x[MAXN + 5], y[MAXN + 5];
double dis[MAXN + 5][MAXN + 5];
double len(int u, int v)
{
    return sqrt((long long)(x[u] - x[v]) * (x[u] - x[v]) +
                (long long)(y[u] - y[v]) * (y[u] - y[v]));
}
double maxLen[MAXN]; // maxLen[i]: 到i号点距离最远的点的距离
signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    // 输入
    cin >> n;
    for (int i = 1; i <= n; i++)
        cin >> x[i] >> y[i];
    for (int i = 1; i <= n; i++)
    {
        for (int j = 1; j <= n; j++)
            dis[i][j] = INF;
        dis[i][i] = 0;
    }
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= n; j++)
        {
            char now;
            cin >> now; // 牧区 i 与牧区 j 的连接关系是 now
            if (now == '1')
            {
                dis[i][j] = len(i, j);
            }
        }
    // floyd
    for (int k = 1; k <= n; k++)
        for (int i = 1; i <= n; i++)
            for (int j = 1; j <= n; j++)
                dis[i][j] = min(dis[i][j],
                                dis[i][k] + dis[k][j]);
    // 预处理距离每个点最远的点的距离
    for (int i = 1; i <= n; i++)
    {
        maxLen[i] = 0;
        for (int j = 1; j <= n; j++)
        {
            // double类型直接比较相等可能会有问题
            // 下面的式子逻辑上等价于 dis[i][j]!=INF
            if (abs(dis[i][j] - INF) > 1e-12)
            {
                maxLen[i] = max(maxLen[i], dis[i][j]);
            }
        }
    }
    // 枚举可能的答案
    double ans = INF;
    for (int i = 1; i <= n; i++)
    {
        for (int j = 1; j <= n; j++)
        {
            // 如果 i，j 属于不同的牧场，尝试加一条边沟通两个牧场
            if (abs(dis[i][j] - INF) < 1e-12)
            {
                if (maxLen[i] + maxLen[j] + len(i, j) < ans)
                {
                    ans = maxLen[i] + maxLen[j] + len(i, j);
                }
            }
        }
    }
    // 有没有可能还不如原来的直径大
    for (int i = 1; i <= n; i++)
        ans = max(ans, maxLen[i]);
    //   输出
    cout << fixed << setprecision(6) << ans << "\n";
    return 0;
}