#include <cstdio>
#include <algorithm>
#include <cmath>#define INPUT_FILE "in.txt"
#define OUTPUT_FILE "out.txt"using namespace std;typedef long long LL;
const int maxn = 400;
const double eps = 1e-10;void setfile() {freopen(INPUT_FILE,"r",stdin);freopen(OUTPUT_FILE,"w",stdout);
}struct Point {double x,y;Point(double x = 0,double y = 0):x(x),y(y) {}
};typedef Point Vector;
Point v[maxn * maxn];   //顶点
Point p[maxn];Vector operator - (Point a,Point b) {return Vector(a.x - b.x,a.y - b.y);
}Vector operator * (Vector a,double p) {return Vector(a.x * p,a.y * p);
}Vector operator + (Vector a,Vector b) {return Vector(a.x + b.x,a.y + b.y);
}double dot(Vector a,Vector b) {return a.x * b.x + a.y * b.y;
}int dcmp(double x) {if(fabs(x) < eps) return 0;return x < 0 ? -1 : 1;
}bool operator < (Point a,Point b) {return a.x < b.x || (a.x == b.x && a.y < b.y);
}double cross(Vector a,Vector b) {return a.x * b.y - a.y * b.x;
}bool have_intersection(Point a1,Point a2,Point b1,Point b2) {double c1 = cross(a2 - a1,b1 - a1),c2 = cross(a2 - a1,b2 - a1);double c3 = cross(b2 - b1,a1 - b1),c4 = cross(b2 - b1,a2 - b1);bool ret = (dcmp(c1) * dcmp(c2) < 0) && (dcmp(c3) * dcmp(c4) < 0);return ret;
}bool on_segment(Point p,Point a,Point b) {return dcmp(cross(p - a,p - b)) == 0 && dcmp(dot(a - p,b - p)) < 0;
}Point get_intersection(Point a1,Point a2,Point b1,Point b2) {Vector v = a2 - a1,w = b2 - b1,u = a1 - b1;double t = cross(w,u) / cross(v,w);return a1 + v * t;
}bool operator == (Point a,Point b) {return dcmp(a.x - b.x) == 0 && dcmp(a.y - b.y) == 0;
}int main() {int n,kase = 1;while(scanf("%d",&n),n) {for(int i = 0;i < n;i++) {scanf("%lf%lf",&v[i].x,&v[i].y);p[i] = v[i];}int cnt_v = n - 1,cnt_f = 0,cnt_e = n - 1;   n--;        //判断一下给定的线段是否有交点for(int i = 0;i < n;i++) {for(int j = i + 1;j < n;j++) {if(have_intersection(p[i],p[i + 1],p[j],p[j + 1])) {//找到交点了就添加到点集里面v[cnt_v++] = get_intersection(p[i],p[i + 1],p[j],p[j + 1]);}}}//特判三点共线的时候的状态sort(v,v + cnt_v);cnt_v = unique(v,v + cnt_v) - v;//处理因为线段交点而新生成的线段for(int i = 0;i < cnt_v;i++) {for(int j = 0;j < n;j++) {if(on_segment(v[i],p[j],p[j + 1])) {cnt_e++;}}}printf("Case %d: There are %d pieces.\n",kase++,2 + cnt_e - cnt_v);}return 0;
}