/*Point operator & (Line A,Line B)
{Point C=A.s;double t=((A.s-B.s)^(B.s-B.e))/((A.s-A.e)^(B.s-B.e));C.x+=(A.e.x-A.s.x)*t; C.y+=(A.e.y-A.s.y)*t;return C;
}半平面交
1.首先规定逆时针给点 判断的时候直接求有向面积 面积为负则是顺时针给的点
2.极角排序 （可以去重 留下靠左的一个）
3.模拟双端队列 判断队列尾部两线的交点和头部两线的交点在新加入线段哪一侧 左侧即合法 直接加入 右侧即不合法 删掉堆尾
4.因为我们 3 过程中始终保持队列数量至少为2 所以最后判断头部和尾部的合法性剩下的边<=2条的话 说明不存在内核
NOTE: 此题特殊 之存在上下左右方向的线  最后决定的线也只有4条 可以直接判断 不用打半平面交所以&交点也不需要考虑两线平行的情况
*/
#include<cmath>
#include<cstdio>
#include<algorithm>
using namespace std;
const int N=150;
const double eps=1e-9;
struct Point
{double x,y;Point(){}Point(double _x,double _y){x=_x,y=_y;}
}p[N];
struct Line
{Point s,e;Line(){}Line(Point _s,Point _e){s=_s,e=_e;}
}l[N],q[N];
int n,T;
Point operator - (Point A,Point B) {return Point(A.x-B.x,A.y-B.y);}
double operator ^ (Point A,Point B) {return A.x*B.y-A.y*B.x;}
Point operator & (Line A,Line B)
{Point C=A.s;double t=((A.s-B.s)^(B.s-B.e))/((A.s-A.e)^(B.s-B.e));C.x+=(A.e.x-A.s.x)*t; C.y+=(A.e.y-A.s.y)*t;return C;
}
bool Isf() // 判断给出的点是否是逆时针的
{double res=0;for(int i=2;i<n;i++) res+=((p[i]-p[1])^(p[i+1]-p[1]));return res>eps;
}
bool check(Line a,Line b,Line c) // 判断 a，b的交点是否在c的左侧包括线上
{Point d=a&b; return ((d-c.s)^(c.e-c.s))<eps;
}
int from(Line a) //  up left down right
{Point A=a.e-a.s;if(A.x==0&&A.y>0) return 1;else if(A.x<0&&A.y==0) return 2;else if(A.x==0&&A.y<0) return 3;else return 4;
}
bool cmp(Line a,Line b)
{int f1=from(a),f2=from(b);if(f1!=f2) return f1<f2;return ((a.s-b.s)^(b.e-b.s))<eps;
}
bool solve()
{sort(l+1,l+n+1,cmp); int top=1;for(int i=2;i<=n;i++) if(from(l[i])!=from(l[top])) l[++top]=l[i];Line up,left,right,down;up=l[1];left=l[2];down=l[3];right=l[4];
//printf("    up : (%.2lf,%.2lf) -> (%.2lf,%.2lf)\n",l[1].s.x,l[1].s.y,l[1].e.x,l[1].e.y);
//printf("    left : (%.2lf,%.2lf) -> (%.2lf,%.2lf)\n",l[2].s.x,l[2].s.y,l[2].e.x,l[2].e.y);
//printf("    down : (%.2lf,%.2lf) -> (%.2lf,%.2lf)\n",l[3].s.x,l[3].s.y,l[3].e.x,l[3].e.y);
//printf("    right : (%.2lf,%.2lf) -> (%.2lf,%.2lf)\n",l[4].s.x,l[4].s.y,l[4].e.x,l[4].e.y);
//printf("    up : (%.2lf,%.2lf)\n",up.x,up.y);
//printf("    left : (%.2lf,%.2lf)\n",left.x,left.y);
//printf("    down : (%.2lf,%.2lf)\n",down.x,down.y);
//printf("    right : (%.2lf,%.2lf)\n",right.x,right.y);if(up.s.x-down.s.x<-eps)return 0;if(left.s.y-right.s.y<-eps) return 0;return 1;
}
int main()
{while(scanf("%d",&n)&&n){T++; for(int i=1;i<=n;i++){double x,y;scanf("%lf%lf",&x,&y);p[i]=Point(x,y);}if(!Isf()) for(int i=1;i<=(n>>1);i++) swap(p[i],p[n-i+1]);p[n+1]=p[1]; for(int i=1;i<=n;i++) l[i]=Line(p[i],p[i+1]);printf("Floor #%d\n",T);if(solve()) puts("Surveillance is possible."); else puts("Surveillance is impossible.");putchar('\n');}return 0;
}
/*
bian8
1 0
1 1
0 1
0 3
1 3
1 2
2 2
2 0dian8
1 1
0 1
0 2
1 2
1 1
2 1
2 0
1 0*/