#include<iostream>
#include<cmath>
#include<string.h>
#include<string>
#include<cstdio>
#include<algorithm>
#include<iomanip>using namespace std;
#define eps 1e-8
#define PI acos(-1.0)//点和向量
struct Point{double x,y;Point(double x=0,double y=0):x(x),y(y){}void out(){cout<<"("<<x<<','<<y<<") ";}
};
typedef Point Vector;
Vector operator+(Vector a,Vector b){return Vector(a.x+b.x,a.y+b.y);}
Vector operator-(Vector a,Vector 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,double p){return Vector(a.x/p,a.y/p);}
bool operator<(const Vector& a,const Vector& b){return a.x<b.x||(a.x==b.x && a.y<b.y);}
int dcmp(double x){if(fabs(x)<eps)return 0;else return x<0 ? -1:1;
}
bool operator==(const Point& a,const Point& b){return dcmp(a.x-b.x)==0 && dcmp(a.y-b.y)==0;
}
double Dot(Vector A,Vector B){return A.x*B.x+A.y*B.y;}//向量点积
double Length(Vector A){return sqrt(Dot(A,A));}//向量模长
double Angle(Vector A,Vector B){return acos(Dot(A,B)/Length(A)/Length(B));}//向量夹角
double Cross(Vector A,Vector B){return A.x*B.y-A.y*B.x;}
double Area2(Point A,Point B,Point C){return Cross(B-A,C-A);}//三角形面积的2倍
//绕起点逆时针旋转rad度
Vector Rotate(Vector A,double rad){          return Vector(A.x*cos(rad)-A.y*sin(rad),A.x*sin(rad)+A.y*cos(rad));
}
double torad(double jiao){return jiao/180*PI;}//角度转弧度
double tojiao(double ang){return ang/PI*180;}//弧度转角度
//单位法向量
Vector Normal(Vector A){double L=Length(A);return Vector(-A.y/L,A.x/L);
}
//点和直线
struct Line{Point P;//直线上任意一点Vector v;//方向向量，他的左边对应的就是半平面double ang;//极角，即从x正半轴旋转到向量v所需的角（弧度）
    Line(){}Line(Point p,Vector v):P(p),v(v){ang=atan2(v.y,v.x);}bool operator<(const Line& L)const {return ang<L.ang;}
};
//计算直线P+tv和Q+tw的交点（计算前必须确保有唯一交点）即：Cross(v,w)非0
Point GetLineIntersection(Point P,Vector v,Point Q,Vector w){Vector u=P-Q;double t=Cross(w,u)/Cross(v,w);return P+v*t;
}
double getx(Point p1,Point p2,Point p3,Point p4)//找到直线和线段相交的横坐标(求两直线交点）
 {double k1=(p2.y-p1.y)/(p2.x-p1.x);double k2=(p4.y-p3.y)/(p4.x-p3.x);double b1=p2.y-k1*p2.x;double b2=p3.y-k2*p3.x;return (b2-b1)/(k1-k2);}
//点到直线距离(dis between point P and line AB)
double DistanceToLine(Point P,Point A,Point B){Vector v1=B-A , v2=P-A;return fabs(Cross(v1,v2))/Length(v1);
}
//dis between point P and segment AB
double DistancetoSegment(Point P,Point A,Point B){if(A==B)return Length(P-A);Vector v1=B-A,v2=P-A,v3=P-B;if(dcmp(Dot(v1,v2))<0)return  Length(v2);else if(dcmp(Dot(v1,v3))>0)return Length(v3);else return fabs(Cross(v1,v2))/Length(v1);
}
//point P on line AB 投影点
Point GetLineProjection(Point P,Point A,Point B){Vector v=B-A;return A+v*(Dot(v,P-A)/Dot(v,v));
}
//线段规范相交（只有一个且不在端点）每条线段两端都在另一条两侧，（叉积符号不同）
bool SegmentProperIntersection(Point a1,Point a2,Point b1,Point b2){a2=(a2-a1)*10000000000+a1;//射线A1A2相交线短B1B2double c1=Cross(a2-a1,b1-a1),c2=Cross(a2-a1,b2-a1),c3=Cross(b2-b1,a1-b1),c4=Cross(b2-b1,a2-b1);return dcmp(c1)*dcmp(c2)<0 && dcmp(c3)*dcmp(c4)<0;
}
//判断点P是否在线段AB上
bool OnSegment(Point p,Point a1,Point a2){return dcmp(Cross(a1-p,a2-p))==0 && dcmp(Dot(a1-p,a2-p))<0;
}
//多边形的面积（可以是非凸多边形）
double PolygonArea(Point* p,int n){double area=0;for(int i=1;i<n-1;i++)area+=Cross(p[i]-p[0],p[i+1]-p[0]);return area/2;
}//点p在有向直线左边，上面不算
bool OnLeft(Line L,Point p){return Cross(L.v,p-L.P)>0;
}double ok(double x,double y,double d,double z){double f=fabs(d*(1/tan(acos(z/y))+1/tan(acos(z/x))))-z;if(fabs(f)<1e-4)return  0;else return f;
}
//计算凸包输入点数组p，个数n，输出点数组ch，返回凸包定点数
//输入不能有重复，完成后输入点顺序被破坏
//如果不希望凸包的边上有输入点，把两个<=改成<
//精度要求高时，建议用dcmp比较
//基于水平的Andrew算法-->1、点排序2、删除重复的然后把前两个放进凸包
//3、从第三个往后当新点在凸包前进左边时继续，否则一次删除最近加入的点，直到新点在左边
int ConVexHull(Point* p,int n,Point*ch){sort(p,p+n);int m=0;for(int i=0;i<n;i++){//下凸包while(m>1 && Cross(ch[m-1]-ch[m-2],p[i]-ch[m-2])<=0)m--;ch[m++]=p[i];}int k=m;for(int i=n-2;i>=0;i--){//上凸包while(m>k && Cross(ch[m-1]-ch[m-2],p[i]-ch[m-2])<=0)m--;ch[m++]=p[i];}if(n>1)m--;return m;
}//******************************************************************************
#define N 21
#define left -10e8
Point up[N],down[N];
int n;double getans()//最值一定过两个顶点一上一下，所以枚举所有点对
 {int i,j,k;double ans=left,right;//二分法double tx,ty;Point ql,qr;for(i=0;i<n;i++)for(j=0;j<n;j++){if(i==j)continue;ql=up[i];qr=down[j];right=left;//left是左边界，非常小的一个值，right就是枚举的过两点的直线最远能达的x的大小for(k=0;k<n;k++)//验证枚举直线是否满足所有点
         {tx=up[k].x;ty=(tx-ql.x)*(qr.y-ql.y)/(qr.x-ql.x)+ql.y;//求出对应x点在枚举的直线上的y值 if(ty>down[k].y&&ty<up[k].y||fabs(ty-down[k].y)<eps||fabs(ty-up[k].y)<eps)//该y值应在上下之间或与上或下重合right=tx;//更新有边界值else//如果该点时不满足
             {if(k)//细节！！！
                 {if(ty<down[k].y)right=getx(ql,qr,down[k-1],down[k]);elseright=getx(ql,qr,up[k-1],up[k]);}break;}}if(right>ans)ans=right;}return ans;}int main(){cout.precision(2);for(;cin>>n&&n;){for(int i=0;i<n;i++){cin>>up[i].x>>up[i].y;down[i].x=up[i].x;down[i].y=up[i].y-1;}double ans=getans();if(ans>up[n-1].x||fabs(ans-up[n-1].x)<eps)cout<<"Through all the pipe.\n";else cout<<fixed<<ans<<'\n';}return 0;
}