#include <set>
#include <map>
#include <deque>
#include <queue>
#include <stack>
#include <cmath>
#include <ctime>
#include <bitset>
#include <cstdio>
#include <string>
#include <vector>
#include <cstdlib>
#include <cstring>
#include <cassert>
#include <iostream>
#include <algorithm>
#include <unordered_map>
using namespace std;
const int INF=0x3f3f3f3f;
typedef long long LL;
typedef pair<LL, LL> pLL;
typedef pair<LL, int> pLi;
typedef pair<int, LL> pil;;
typedef pair<int, int> pii;
typedef unsigned long long uLL;
//这道题没有考虑加速度，且是在塔外面往塔上跳
//长时间未敲代码，double的输入输出都不会了
//既然一定跳到圆锥上,那么可以假设这点是（x,y,z）,然后根据圆锥的截面建立方程x*x+y*y=r1*r1
double x,y,z,vx,vy,vz,r,h;
bool check(double t)
{
double x1=x+vx*t;
double y1=y+vy*t;
double z1=z+vz*t;
//printf("%f\n",z1);
if(z1<0||z1>h)return false;
return (x1*x1+y1*y1-(1-z1/h)*(1-z1/h)*r*r)<=1e-8;
}
int main()
{
int t;
scanf("%d",&t);
int kase=0;
while(t--)
{
scanf("%lf%lf%lf%lf%lf%lf%lf%lf",&r,&h,&x,&y,&z,&vx,&vy,&vz);
// printf("%f %f %f %f %f %f %f %f\n",r,h,x,y,z,vx,vy,vz);
double a=vz*vz*r*r/(h*h)-vx*vx-vy*vy;
double b=2*z*vz*r*r/(h*h)-2*vz*r*r/h-2*x*vx-2*y*vy;
double c=r*r*((h-z)/h)*((h-z)/h)-x*x-y*y;
//printf("%f %f %f\n",a,b,c);
double t1=(-b+sqrt(b*b-4*a*c))/(2*a);
double t2=(-b-sqrt(b*b-4*a*c))/(2*a);
//
// printf("%f ==== %f\n",t1,t2);
double ans=INF;
if(check(t1))ans=min(ans,t1);
if(check(t2))ans=min(ans,t2);
printf("Case %d: %.10f\n",++kase,ans);
}
return 0;
}