/*

  HDU5130 Signal Interference

  http://acm.hdu.edu.cn/showproblem.php?pid=5130

  计算几何 圆与多边形面积交

  *

  */

 #include <cstdio>

 #include <algorithm>

 #include <cmath>

 using namespace std;

 const double Pi=acos(-1.0);

 const double eps = 1e-;

 const int Nmax=;

 double k;

 int sgn(double x)

 {

     if(x<-eps)

         return -;

     if(x>eps)

         return ;

     return ;

 }

 double Abs(double x)

 {

     if(sgn(x)<)

         x=-x;

     else if(sgn(x)==)

         x=0.0;

     return x;

 }

 double sqr(double x)

 {

     return x*x;

 }

 double Sqrt(double x)

 {

     return max(0.0,sqrt(x));

 }

 struct Pt

 {

     double x,y;

     Pt() { }

     Pt(double x, double y) : x(x), y(y) { }

     Pt operator - (const Pt &b)

     {

         return Pt(x-b.x,y-b.y);

     }

     Pt operator + (const Pt &b)

     {

         return Pt(x+b.x,y+b.y);

     }

     friend Pt operator * (double a,Pt p)

     {

         return Pt(p.x*a,p.y*a);

     }

     friend Pt operator * (Pt p,double a)

     {

         return Pt(p.x*a,p.y*a);

     }

     friend Pt operator / (Pt p,double a)

     {

         return Pt(p.x/a,p.y/a);

     }

     double norm()

     {

         return Sqrt(x*x + y*y);

     }

     double len()

     {

         return norm();

     }

     void print()

     {

         printf("(%f, %f)\n", x, y);

     }

 };

 int n;

 Pt pl[Nmax];

 Pt pr,A,B,p0,p1;

 double dist (Pt a, Pt b)

 {

     return (a-b).norm();

 }

 double dot(Pt a,Pt b)//点乘

 {

     return a.x*b.x+a.y*b.y;

 }

 double det(Pt a,Pt b)

 {

     return a.x*b.y-a.y*b.x;

 }

 Pt rotate(Pt p,double a)//P点绕原点逆时针旋转a弧度

 {

     return Pt(p.x*cos(a)-p.y*sin(a),p.x*sin(a)+p.y*cos(a));

 }

 struct Sg//线段

 {

     Pt s, t;

     Sg() { }

     Sg(Pt s, Pt t) : s(s), t(t) { }

     Sg(double a, double b, double c, double d) : s(a, b), t(c, d) { }

 };

 bool PtOnSegment(Pt p, Pt a, Pt b) //p是否在线段ab上，把<=改成<就能实现不含线段端点的点在线段上的判断。

 {

     return !sgn(det(p-a, b-a)) && sgn(dot(p-a, p-b)) <= ;

 }

 bool PtOnLine(Pt p, Pt a, Pt b) //p是否在直线ab上

 {

     return !sgn(det(p-a, b-a));

 }

 Pt PtLineProj(Pt s, Pt t, Pt p) //p到直线st的投影

 {

     double r = dot(p-s, t-s) / (t - s).norm();

     return s + (t - s) * r;

 }

 bool parallel(Pt a, Pt b, Pt s, Pt t)

 {

     return !sgn(det(a-b, s-t));

 }

 Pt triangleMassCenter(Pt a, Pt b, Pt c)

 {

     return (a+b+c) / 3.0;

 }

 double polygon_area(Pt poly[],int n)

 {

     double ans=0.0;

     if(n<)

         return ans;

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

         ans+=det(poly[i],poly[i%n+]);

     return ans*0.5;

 }

 struct Circle

 {

     Pt c;

     double r;

     Circle(Pt _c,double _r)

     {

         c=_c;

         r=_r;

     }

 };

 bool PointInCircle(Circle a,Pt p)

 {

     Pt c=a.c;

     return sgn( (p-c).norm()-a.r )<;

 }

 bool PointOnCircle(Circle a,Pt p)

 {

     Pt c=a.c;

     return sgn( (p-c).norm()-a.r )==;

 }

 bool PointIOCircle(Circle a,Pt p)

 {

     Pt c=a.c;

     return sgn( (p-c).norm()-a.r )<=;

 }

 void circle_cross_line(Circle c,Pt a, Pt b, Pt ans[], int &num)

 {

     double x1=a.x,y1=a.y,x2=b.x,y2=b.y;

     double dx=x2-x1,dy=y2-y1;

     double tmpx=x1-c.c.x,tmpy=y1-c.c.y;

     double A=dx*dx+dy*dy;

     double B=2.0*( dx*tmpx+dy*tmpy );

     double C=tmpx*tmpx+tmpy*tmpy-c.r*c.r;

     double delta=B*B-4.0*A*C;

     num=;

     if(sgn(delta)<)

         return;

     double t1=(-B-Sqrt(delta))/(2.0*A);

     double t2=(-B+Sqrt(delta))/(2.0*A);

     if(sgn(t1-1.0)<= && sgn(t1)>=)

         ans[++num]=Pt(x1+t1*dx,y1+t1*dy);

     if(sgn(delta)==)

         return;

     if(sgn(t2-1.0)<= && sgn(t2)>=)

         ans[++num]=Pt(x1+t2*dx,y1+t2*dy);

 }   

 double sector_area(Circle c,Pt a,Pt b)

 {

     a=a-c.c,b=b-c.c;

     double theta = atan2(a.y, a.x) - atan2(b.y, b.x);

     while (sgn(theta) <= ) theta += 2.0*Pi;

     while (sgn(theta-2.0*Pi)>) theta -= 2.0*Pi;

     theta = min(theta, 2.0*Pi - theta);

     return c.r*c.r*theta*0.5;

 }

 double CirclePolyArea(Circle c,Pt poly[],int n)

 {

     double ans=0.0;

     Pt p[];

     int num;

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

     {

         Pt a=poly[i],b=poly[i%n+];

         int ina=PointInCircle(c,a);

         int inb=PointInCircle(c,b);

         Pt da=a-c.c,db=b-c.c;

         int s=sgn(det(da,db));

         double part=0.0;

         if(ina)

         {

             if(inb)

             {

                 part=Abs(det(da,db))*0.5;

             }

             else

             {

                 circle_cross_line(c,a,b,p,num);

                 part=sector_area(c,p[],b)+Abs(det(da,p[]-c.c))*0.5;

             }

         }

         else

         {

             if(inb)

             {

                 circle_cross_line(c,a,b,p,num);

                 part=sector_area(c,p[],a)+Abs(det(db,p[]-c.c))*0.5;

             }

             else

             {

                 circle_cross_line(c,a,b,p,num);

                 if(num==)

                 {

                     part=sector_area(c,a,p[])+sector_area(c,b,p[])+Abs(det(p[]-c.c,p[]-c.c))*0.5;

                 }

                 else

                 {

                     part=sector_area(c,a,b);

                 }

             }

         }

         if(s!=)

             ans+=1.0*s*part;

     }

     return ans;

 }

 int main()

 {

    int t=;

     while(scanf("%d%lf",&n,&k)==)

     {

         t++;

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

             scanf("%lf%lf",&pl[i].x,&pl[i].y);

         scanf("%lf%lf%lf%lf",&A.x,&A.y,&B.x,&B.y);

         double r;

         double E=(2.0*B.x-2.0*sqr(k)*A.x)/(1.0-sqr(k));

         double F=(2.0*B.y-2.0*sqr(k)*A.y)/(1.0-sqr(k));

         double G=(sqr(k*A.x)+sqr(k*A.y)-sqr(B.x)-sqr(B.y))/(1.0-sqr(k));

         pr.x=E/2.0,pr.y=F/2.0;

         r=Sqrt(G+sqr(E)/4.0+sqr(F)/4.0);

         Circle C(pr,r);

         double ans=Abs( CirclePolyArea(C,pl,n) );

         printf("Case %d: ",t);

         printf("%.10f\n",ans+eps);

     }

     return ;

 }