/*

     最短路：Floyd算法模板题

 */

 #include <cstdio>

 #include <iostream>

 #include <algorithm>

 #include <cmath>

 #include <cstring>

 #include <string>

 #include <vector>

 using namespace std;

 const int MAXN =  + ;

 const int INF = 0x3f3f3f3f;

 double d[MAXN][MAXN];

 int used[MAXN];

 struct NODE

 {

     int x, y;

 }node[MAXN];

 double dis(int j, int i)

 {

     return sqrt ((node[j].x - node[i].x) * (node[j].x - node[i].x) + (node[j].y - node[i].y) * (node[j].y - node[i].y));

 }

 void Floyd_Warshall(int n)

 {

     for (int k=; k<=n; ++k)

     {

         for (int i=; i<=n-; ++i)

         {

             for (int j=i+; j<=n; ++j)

             {

                 if (d[i][k] < d[i][j] && d[k][j] < d[i][j])

                 {

                     if (d[i][k] < d[k][j])

                         d[i][j] = d[j][i] = d[k][j];

                     else

                         d[i][j] = d[j][i] = d[i][k];

                 }

             }

         }

     }

     printf ("Frog Distance = %.3f\n", d[][]);

 }

 int main(void)      //POJ 2253  Frogger

 {

     //freopen ("D.in", "r", stdin);

     int n;

     int cnt = ;

     int first = ;

     while (~scanf ("%d", &n) && n)

     {

         for (int i=; i<=n; ++i)

         {

             scanf ("%d%d", &node[i].x, &node[i].y);

         }

         for (int i=; i<=n-; ++i)

         {

             for (int j=i+; j<=n; ++j)

             {

                 d[i][j] = d[j][i] = dis (i, j);

             }

         }

         printf ("Scenario #%d\n", ++cnt);

         Floyd_Warshall (n);

         puts ("");

     }

     return ;

 }

 /*

 Scenario #1

 Frog Distance = 5.000

 Scenario #2

 Frog Distance = 1.414

 */