1 /*
 2 几何+直线相交+求交点
 3 */
 4 #include<stdio.h>
 5 #include<string.h>
 6 #include<stdlib.h>
 7 #include<algorithm>
 8 using namespace std;
 9 const int maxn = 24;
10 const double inf= 99999999.0;
11 const double eps = 1e-8;
12 struct Point{
13     double x,y;
14 }up[ maxn ],down[ maxn ];
15 int sig( double d ){
16     if( d>eps ) return 1;
17     if( d<-eps ) return -1;
18     return 0;
19 }
20 double xmult( Point a,Point b,Point c ){
21     return ( b.x-a.x )*( c.y-a.y )-( b.y-a.y )*( c.x-a.x );
22 }
23 bool check( Point a,Point b,Point c,Point d ){
24     return (( sig(xmult(a,b,c)) )*( sig(xmult(a,b,d)) )<=0);
25 }
26 double Intersection( Point a,Point b,Point c,Point d ){
27     double s1 = xmult( a,b,c );
28     double s2 = xmult( a,b,d );
29     int t1,t2;
30     t1 = sig( s1 ),t2 = sig( s2 );
31     if( t1*t2<0 ){
32         return ( c.x*s2-d.x*s1 )/(s2-s1);//return x 坐标
33         //(c.y*s2-d.y*s1)/(s2-s1)
34     }
35     if( t1*t2==0 ){
36         if( t1==0 ) return c.x;
37         if( t2==0 ) return d.x;
38     }
39     return -inf;
40 }
41 int main(){
42     int n;
43     while( scanf("%d",&n)==1,n ){
44         for( int i=1;i<=n;i++ ){
45             scanf("%lf%lf",&up[ i ].x,&up[ i ].y);
46             down[ i ].x = up[ i ].x;
47             down[ i ].y = up[ i ].y-1.0;
48         }
49         double ans = -inf;
50         bool flag = false;//是否能贯通两端
51         for( int i=1;i<=n;i++ ){
52             for( int j=1;j<=n;j++ ){
53                 if( i!=j ){
54                     int k;
55                     for( k=1;k<=n;k++ ){
56                         if( check( up[i],down[j],up[k],down[k] )==true ){}//与垂直的线段相交true
57                         else break;
58                     }
59                     if( k>n ){
60                         flag = true;
61                         break;
62                     }
63                     else if( k>max(i,j) ){
64                         double pos;
65                         pos = Intersection( up[i],down[j],up[k],up[k-1] );
66                         if( pos>ans ) ans = pos;
67                         pos = Intersection( up[i],down[j],down[k],down[k-1] );
68                         if( pos>ans ) ans = pos;
69                         //因为不知道ij是和上管壁相交还是下管壁相交
70                     }
71                 }
72             }
73             if( flag==true ) break;
74         }
75         if( flag == true ) printf("Through all the pipe.\n");
76         else printf("%.2lf\n",ans);
77     }
78     return 0;
79 }