/*
題意:給出一個無向圖,求一條1~2的路徑使得路徑上的最大邊權最小.

分析:dijkstra變形,將更新距離的過程改為取最大值即可.
*/
#include<stdio.h>
#include<math.h>
#include<string.h>
#include<algorithm>
using namespace std;

const double DNF = 2000;
const int maxn = 210;

double w[maxn][maxn];
double d[maxn];
int vis[maxn];
int n;
double ans;

struct Point{
    double x,y;
}p[maxn];

double dist(Point a, Point b)
{
    return sqrt((a.x-b.x)*(a.x-b.x) + (a.y-b.y)*(a.y-b.y));
}

void Dijkstra()
{
    for(int i = 1; i <= n; i++) d[i] = DNF;
    d[1] = 0;
    memset(vis,0,sizeof(vis));

    for(int i = 1; i <= n; i++)
    {
        int x;
        double m = DNF;
        for(int y = 1; y <= n; y++) if(!vis[y] && d[y] <= m) m = d[x=y];
        vis[x] = 1;
        if(ans < d[x] && d[x]!= DNF) // ans 是這條路徑上的最大權
        {
            ans = d[x];
        }
        if(x == 2) return; //走到目的地即可
        for(int y = 1; y <= n; y++) if(!vis[y])
            d[y] = min(d[y], w[x][y]); //更新未接入的點的dist
    }

}

int main()
{
    int kcase = 0;
    while(scanf("%d", &n) != EOF)
    {
        if(n == 0) break;
        for(int i = 1; i <= n; i++)
        {
            scanf("%lf%lf", &p[i].x, &p[i].y);
        }

        for(int i = 1; i <= n; i++)
        {
            w[i][i] = 0;
            for(int j = i+1; j <= n; j++)
            {
                w[i][j] = dist(p[i],p[j]);
                w[j][i] = dist(p[i],p[j]);
            }
        }

        ans = 0;
        Dijkstra();
        printf("Scenario #%d\n", ++kcase);
        printf("Frog Distance = %.3lf\n\n", ans);
    }
}