/*题目大意：给出平面上一些点，问一个半径为1的圆最多可以覆盖几个点题解：我们对于每个点画半径为1的圆，那么在两圆交弧上的点所画的圆，一定可以覆盖这两个点我们对于每个点计算出其和其它点的交弧，对这些交弧计算起末位置对于圆心的极角，对这些我们进行扫描线操作，统计最大交集数量就是答案。
*/
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <cstring>
using namespace std;
double EPS=1e-10;
double add(double a,double b){if(abs(a+b)<EPS*(abs(a)+abs(b)))return 0;return a+b;
}
const int MAX_N=310;
struct P{double x,y;P(){}P(double x,double y):x(x),y(y){}P operator + (P p){return P(add(x,p.x),add(y,p.y));}P operator - (P p){return P(add(x,-p.x),add(y,-p.y));}P operator * (double d){return P(x*d,y*d);}double dot(P p){return add(x*p.x,y*p.y);} //点积double det(P p){return add(x*p.y,-y*p.x);}  //叉积
}ps[MAX_N];
double dist(P p,P q){return sqrt((p-q).dot(p-q));}
struct PolarAngle{double angle;bool flag;
}as[MAX_N];
bool cmp_a(PolarAngle a,PolarAngle b){return a.angle<b.angle;
}
int solve(int n,double r){int ans=1;for(int i=0;i<n;i++){int m=0; double d;for(int j=0;j<n;j++){if(i!=j&&(d=dist(ps[i],ps[j]))<=2*r){double phi=acos(d/2);double theta=atan2(ps[j].y-ps[i].y,ps[j].x-ps[i].x);as[m].angle=theta-phi,as[m++].flag=1;as[m].angle=theta+phi,as[m++].flag=0;}}sort(as,as+m,cmp_a);for(int sum=1,j=0;j<m;j++){if(as[j].flag)sum++;else sum--;ans=max(ans,sum);}}return ans;
}
int N;
int main(){while(scanf("%d",&N),N){for(int i=0;i<N;i++)scanf("%lf%lf",&ps[i].x,&ps[i].y);printf("%d\n",solve(N,1.0));}return 0;
}

/*给出一些圆的位置和半径以及叠放次序，问能看到的有几个圆
*/
#include <cstring>
#include <algorithm>
#include <cmath>
#include <complex>
#include <vector>
#include <cstdio>
using namespace std;
typedef complex<double> P;
#define M_PI 3.14159265358979323846
const double EPS=4E-13;
double Tran(double r){while(r<0.0)r+=2*M_PI;while(r>=2*M_PI)r-=2*M_PI;return r;
}
int hit_test(P p,vector<P>&points,vector<double> &rs){for(int i=rs.size()-1;i>=0;i--){if(abs(points[i]-p)<rs[i])return i;}return -1;
}
int n;
int main(){while(scanf("%d",&n),n){vector<P> points;vector<double> rs;for(int i=0;i<n;i++){double x,y,r;scanf("%lf%lf%lf",&x,&y,&r);points.push_back(P(x,y));rs.push_back(r);}vector<bool> visible(n,false);for(int i=0;i<n;i++){vector<double> rads;rads.push_back(0.0);rads.push_back(2.0*M_PI);for(int j=0;j<n;j++){double a=rs[i],b=abs(points[j]-points[i]),c=rs[j];if(a+b<c||a+c<b||b+c<a)continue;double d=arg(points[j]-points[i]);double e=acos((a*a+b*b-c*c)/(2*a*b));rads.push_back(Tran(d+e));rads.push_back(Tran(d-e));}sort(rads.begin(),rads.end());for(int j=0;j<rads.size()-1;j++){double rad=(rads[j+1]+rads[j])/2.0;for(int k=-1;k<=1;k+=2){int t=hit_test(P(points[i].real()+(rs[i]+EPS*k)*cos(rad),points[i].imag()+(rs[i]+EPS*k)*sin(rad)),points,rs);if(t!=-1)visible[t]=true;}}}printf("%d\n",count(visible.begin(),visible.end(),true));}return 0;
}

/*题目大意：给出一些圆，求一条直线最多与几个圆相切题解：我们枚举任意两个圆的切线，然后计算与这条切线相切的圆的数目即可。
*/
#include <iostream>
#include <vector>
#include <complex>
#include <cmath>
#include <algorithm>
using namespace std;
typedef complex<double> P;
const double PI=acos(-1);
const double EPS=1e-12;
int cmp(double a,double b){const double diff=a-b;if(fabs(diff)<EPS)return 0;else if(diff<0)return -1;else return 1;
}
inline double dot(const P &a, const P &b){return a.real()*b.real()+a.imag()*b.imag();
}
inline double cross(const P &a, const P &b){return a.real()*b.imag()-b.real()*a.imag();
}
struct line{P a,b;line(){}line(const P &p,const P &q):a(p),b(q){}// 是否平行inline bool parallel(const line &ln) const{return abs(cross(ln.b-ln.a,b-a))<EPS; //平行叉乘得到向量的模是0，也就是sin(theta)=0<->theta=0}// 是否相交inline bool intersects(const line &ln) const{return !parallel(ln);}// 求交点inline P intersection(const line &ln) const{const P x=b-a;const P y=ln.b-ln.a;return a+x*(cross(y,ln.a-a))/cross(y,x);}// 点到直线的距离inline double distance(const P &p) const{return abs(cross(p-a,b-a))/abs(b-a);}// 求垂足坐标inline P perpendicular(const P &p) const{const double t=dot(p-a,a-b)/dot(b-a,b-a);return a+t*(a-b);}
};
struct circle{P o;double r;circle(){}circle(const P &p,double x):o(p),r(x){}// 通过点 p 的两条切线pair<P,P> tangent(const P &p)const{const double L=abs(o-p);const double M=sqrt(L*L-r*r);const double theta=asin(r/L);const P v=(o-p)/L;return make_pair(p+M*(v*polar(1.0,theta)),p+M*(v*polar(1.0,-theta)));}// 两个半径相等圆的两条平行外切线pair<line,line> outer_tangent_parallel(const circle &c) const{const P d=o-c.o;const P v=d*P(0,1)*r/abs(d);return make_pair(line(o+v,c.o+v),line(o-v,c.o-v));}// 两个圆外切线pair<line,line> outer_tangent(const circle &c) const{if(cmp(r,c.r)==0)return outer_tangent_parallel(c);if(r>c.r)return c.outer_tangent(*this);const P d=o-c.o;const double fact=c.r/r-1;const P base=c.o+d+d/fact;const pair<P, P> t=tangent(base);return make_pair(line(base,t.first),line(base,t.second));}// 内切线pair<line,line> inner_tangent(const circle &c) const{if(r>c.r)return c.inner_tangent(*this);const P d=c.o-o;const double fact=c.r/r+1;const P base=o+d/fact;const pair<P,P> t=tangent(base);return make_pair(line(base,t.first),line(base,t.second));}// 是否相交inline bool intersects(const circle &c) const{return !contains(c)&&!c.contains(*this)&&cmp(abs(o-c.o),r+c.r)<=0;}// 是否相离inline bool independent(const circle &c) const{return cmp(abs(o-c.o),r+c.r)>0;}// 两个圆的交点pair<P,P> intersection(const circle &c) const{const double d=abs(o-c.o);const double cos_=(d*d+r*r-c.r*c.r)/(2*d);const double sin_=sqrt(r*r-cos_*cos_);const P e=(c.o-o)/d;return make_pair(o+e*P(cos_,sin_),o+e* P(cos_,-sin_));}// 是否包含圆cinline bool contains(const circle &c) const{return cmp(abs(o-c.o)+c.r,r)<0;}// 是否相交inline bool intersects(const line &ln) const{return cmp(abs(ln.distance(o)),r)<=0;}// 圆心到直线的距离inline double distance(const line &ln) const{return abs(ln.distance(o));}// 圆与直线的交点pair<P,P> intersection(const line &ln) const{const P h=ln.perpendicular(o);const double d=abs(h-o);P ab=ln.b-ln.a;ab/=abs(ab);const double l=sqrt(r*r-d*d);return make_pair(h+l*ab,h-l*ab);}
};
void enum_event(const circle &c1,const circle &c2,vector<line> &lines){if(c1.independent(c2)){pair<line,line> outer=c1.outer_tangent(c2);lines.push_back(outer.first);lines.push_back(outer.second);pair<line,line> inner = c1.inner_tangent(c2);lines.push_back(inner.first);lines.push_back(inner.second);}else if (c1.intersects(c2)){pair<line,line> outer=c1.outer_tangent(c2);lines.push_back(outer.first);lines.push_back(outer.second);pair<P,P> inter=c1.intersection(c2);lines.push_back(line(inter.first,inter.second));   // 此时内切线不存在，使用交点形成的线代替}
}
bool solve(){int N;scanf("%d",&N);if(!N)return false;vector<pair<circle,circle> > jewels;vector<line> lines;for(int i=0;i<N;i++){double x,y,r,m;scanf("%lf%lf%lf%lf",&x,&y,&r,&m);const P center(x,y);pair<circle,circle> jewel=make_pair(circle(center,r),circle(center,r+m));for(const auto &other:jewels){enum_event(jewel.first,other.first,lines);enum_event(jewel.first,other.second,lines);enum_event(jewel.second,other.first,lines);enum_event(jewel.second,other.second,lines);}jewels.push_back(jewel);}int ans=1;for(auto &l:lines){int cnt=count_if(jewels.begin(),jewels.end(),[&](const pair<circle,circle> &j){ // [&] 按引用捕获在lambda表达式所在函数的函数体中提及的全部自动储存持续性变量return cmp(j.first.r, j.first.distance(l))<=0&&cmp(j.second.r,j.second.distance(l))>=0;    // 在磁力圆范围内且不在本体范围内});ans=max(ans, cnt);}printf("%d\n",ans);return 1;
}
int main(){while(solve());return 0;
}

/*题目大意：给出一些矩形，没有相交和包含的情况，只有相切的情况问有多少个矩形没有相切或者边角重叠题解：我们将所有的与x轴平行的线段和与y周平行的线段分开处理，判断是否出现重合对重合的两个矩形进行标识，最后没有被标识过的矩形数目就是答案。
*/
#include <cstdio>
#include <vector>
#include <algorithm>
#include <cstring>
using namespace std;
const int N=30010;
struct data{int id,d,x,y;data(){}; data(int _d,int _x,int _y,int _id):d(_d),x(_x),y(_y),id(_id){}
};
vector<data> sx,sy;
bool vis[N];
bool cmp(data a,data b){if(a.d!=b.d)return a.d<b.d;if(a.x!=b.x)return a.x<b.x;return a.y<b.y;
}
int n,a,b,c,d;
void solve(){sx.clear();sy.clear();memset(vis,0,sizeof(vis));for(int i=0;i<n;i++){scanf("%d%d%d%d",&a,&b,&c,&d);sy.push_back(data(b,a,c,i));sy.push_back(data(d,a,c,i));  sx.push_back(data(a,b,d,i));  sx.push_back(data(c,b,d,i));}sort(sx.begin(),sx.end(),cmp);sort(sy.begin(),sy.end(),cmp);int t=sy[0].y;for(int i=1;i<sy.size();i++){if(sy[i-1].d==sy[i].d){if(t>=sy[i].x){vis[sy[i].id]=vis[sy[i-1].id]=1;}}else t=sy[i].y;t=max(sy[i].y,t);}t=sx[0].y;for(int i=1;i<sx.size();i++){if(sx[i-1].d==sx[i].d){if(t>=sx[i].x){vis[sx[i].id]=vis[sx[i-1].id]=1;}}else t=sx[i].y;t=max(sx[i].y,t);}int ans=0;for(int i=0;i<n;i++)if(!vis[i])ans++;printf("%d\n",ans);
}
int main(){while(~scanf("%d",&n))solve();return 0;
}

/*题目大意：给出一些点，每个点只能向外引出一条平行X轴，和Y轴的边，问能否构成一个闭多边形，如果能，返回多边形的总边长，否则返回-1题解：我们发现对于每一行或者每一列都必须有偶数个点，且两两之间相邻才能满足条件所以我们将其连线之后判断是否可以构成一个封闭图形，同时还需要判断这些线是否会相交，如果相交即不成立
*/
#include <cstdio>
#include <algorithm>
using namespace std;
const int N=100010;
struct Point{int x,y,id;}p[N];
struct Line{int d,x,y;Line(){}Line(int _d,int _x,int _y):d(_d),x(_x),y(_y){}
}l[N];
int cmp_x(Point a,Point b){if(a.x==b.x)return a.y<b.y;return a.x<b.x;
}
int cmp_y(Point a,Point b){if(a.y==b.y)return a.x<b.x;return a.y<b.y;
}
int con[N][2],n,ln,T;
int Check(Point a,Point b){int y=a.y,x1=a.x,x2=b.x;for(int i=0;i<ln;i++){if(x1<l[i].d&&x2>l[i].d&&l[i].x<y&&l[i].y>y)return 1;}return 0;
}
int main(){scanf("%d",&T);while(T--){scanf("%d",&n);for(int i=0;i<n;i++){scanf("%d%d",&p[i].x,&p[i].y);p[i].id=i;}int s=0,cnt=1,flag=0;ln=0;sort(p,p+n,cmp_x);for(int i=1;i<n&&!flag;i++){if(p[i].x!=p[i-1].x){if(cnt&1)flag=1;cnt=1;}else{cnt++;if((cnt&1)==0){s+=p[i].y-p[i-1].y;con[p[i].id][0]=p[i-1].id;con[p[i-1].id][0]=p[i].id;l[ln++]=Line(p[i].x,p[i-1].y,p[i].y);}}}sort(p,p+n,cmp_y);cnt=1;for(int i=1;i<n&&!flag;i++){if(p[i].y!=p[i-1].y){if(cnt&1)flag=1;cnt=1;}else{cnt++;if((cnt&1)==0){s+=p[i].x-p[i-1].x;con[p[i].id][1]=p[i-1].id;con[p[i-1].id][1]=p[i].id;if(Check(p[i-1],p[i]))flag=1;}}}int t=1,x=0,c=0;for(;;){x=con[x][t];t^=1; c++;if(x==0||flag)break;}if(c!=n)flag=1;if(flag)puts("-1");else printf("%d\n",s);}return 0;
}

/*题目大意：给出一些点的二维坐标和权值，求用一个长H，宽W的矩形能框住的最大权值之和，在矩形边缘的点不计算在内题解：我们计算能扫到这个点的区间范围，将其拆分为两条平行于y轴的左闭右开的直线，为方便边界处理，我们将坐标扩大两倍，之后我们按照x轴对这些线段进行扫描统计出现的最大值即可。
*/
#include <cstdio>
#include <algorithm>
#include <utility>
using namespace std;
typedef long long LL;
const int N=10010;
LL xs[N],ys[N],X[N<<1],Y[N<<1];
int cs[N],tag[N<<3],T[N<<3];
pair<pair<int,int>,pair<int,int> >event[N<<1];
void update(int L,int R,int v,int x,int l,int r){if(L<=l&&r<=R){T[x]+=v;tag[x]+=v;return;}int mid=(l+r)>>1;if(L<=mid)update(L,R,v,x<<1,l,mid);if(mid<R)update(L,R,v,x<<1|1,mid+1,r);T[x]=max(T[x<<1],T[x<<1|1])+tag[x];
}
int n,W,H;
void solve(){for(int i=0;i<n;i++){scanf("%lld%lld%d",xs+i,ys+i,cs+i);xs[i]<<=1; ys[i]<<=1;}for(int i=0;i<n;i++){X[i<<1]=xs[i]-W; X[i<<1|1]=xs[i]+W;Y[i<<1]=ys[i]-H; Y[i<<1|1]=ys[i]-1+H;}sort(X,X+n*2);sort(Y,Y+n*2);for(int i=0;i<n;i++){event[i<<1]=make_pair(make_pair(lower_bound(X,X+n*2,xs[i]-W)-X,cs[i]),make_pair(lower_bound(Y,Y+n*2,ys[i]-H)-Y,lower_bound(Y,Y+n*2,ys[i]+H-1)-Y));event[i<<1|1]=make_pair(make_pair(lower_bound(X,X+n*2,xs[i]+W)-X,-cs[i]),make_pair(lower_bound(Y,Y+n*2,ys[i]-H)-Y,lower_bound(Y,Y+n*2,ys[i]+H-1)-Y));}sort(event,event+n*2);int ans=0;for(int i=0;i<n*2;i++){   update(event[i].second.first,event[i].second.second,event[i].first.second,1,0,n*2);ans=max(ans,T[1]);}printf("%d\n",ans);
}
int main(){while(~scanf("%d%d%d",&n,&W,&H))solve();return 0;
}

/*题目大意：给出一个城堡，要求求出距城堡距离大于L的地方建围墙将城堡围起来求所要围墙的长度题解：画图易得答案为凸包的周长加一个圆的周长。
*/
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <vector>
using namespace std;
double EPS=1e-10;
const double PI=acos(-1.0);
double add(double a,double b){if(abs(a+b)<EPS*(abs(a)+abs(b)))return 0;return a+b;
}
struct P{double x,y;P(){}P(double x,double y):x(x),y(y){}P operator + (P p){return P(add(x,p.x),add(y,p.y));}P operator - (P p){return P(add(x,-p.x),add(y,-p.y));}P operator * (double d){return P(x*d,y*d);}double dot(P p){return add(x*p.x,y*p.y);} //点积double det(P p){return add(x*p.y,-y*p.x);}  //叉积
};
bool cmp_x(const P& p,const P& q){if(p.x!=q.x)return p.x<q.x;return p.y<q.y;
}
vector<P> convex_hull(P* ps,int n){sort(ps,ps+n,cmp_x);int k=0;vector<P> qs(n*2);for(int i=0;i<n;i++){while((k>1)&&(qs[k-1]-qs[k-2]).det(ps[i]-qs[k-1])<=0)k--;qs[k++]=ps[i];}for(int i=n-2,t=k;i>=0;i--){while(k>t&&(qs[k-1]-qs[k-2]).det(ps[i]-qs[k-1])<=0)k--;qs[k++]=ps[i];}qs.resize(k-1);return qs;
}
double dist(P p,P q){return sqrt((p-q).dot(p-q));}
const int MAX_N=1000;
int N,L;
P ps[MAX_N];
vector<P> con;
void solve(){for(int i=0;i<N;i++)scanf("%lf%lf",&ps[i].x,&ps[i].y);con=convex_hull(ps,N);double res=0;for(int i=0;i<con.size()-1;i++)res+=dist(con[i],con[i+1]);res+=dist(con[0],con[con.size()-1]);res+=2*PI*L;printf("%d\n",(int)(res+0.5));
}
int main(){while(~scanf("%d%d",&N,&L))solve();return 0;
}

/*题目大意：给出一些点，表示一些屋子，这些屋子共同组成了村庄，现在要建一些高速公路问是否经过了村庄。题解：这些屋子的关键点一定在凸包上，所以我们只要求出凸包，判断是否和线相交即可我们求出与高速公路相近和近似相反的向量，判断连线是否与这条公路相交即可。
*/
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <vector>
using namespace std;
double EPS=1e-10;
const double PI=acos(-1.0);
double add(double a,double b){if(abs(a+b)<EPS*(abs(a)+abs(b)))return 0;return a+b;
}
struct P{double x,y;P(){}P(double x,double y):x(x),y(y){}P operator + (P p){return P(add(x,p.x),add(y,p.y));}P operator - (P p){return P(add(x,-p.x),add(y,-p.y));}P operator * (double d){return P(x*d,y*d);}double dot(P p){return add(x*p.x,y*p.y);} //点积double det(P p){return add(x*p.y,-y*p.x);}  //叉积
};
bool cmp_x(const P& p,const P& q){if(p.x!=q.x)return p.x<q.x;return p.y<q.y;
}
vector<P> convex_hull(P* ps,int n){sort(ps,ps+n,cmp_x);int k=0;vector<P> qs(n*2);for(int i=0;i<n;i++){while((k>1)&&(qs[k-1]-qs[k-2]).det(ps[i]-qs[k-1])<=0)k--;qs[k++]=ps[i];}for(int i=n-2,t=k;i>=0;i--){while(k>t&&(qs[k-1]-qs[k-2]).det(ps[i]-qs[k-1])<=0)k--;qs[k++]=ps[i];}qs.resize(k-1);return qs;
}
double dist(P p,P q){return sqrt((p-q).dot(p-q));}
double normalize(double r){if(r<-PI/2.0+EPS)r+=PI*2;return r;
}
double atan2(const P& p){return normalize(atan2(p.y, p.x));
}
bool double_cmp(double a,double b){return a+EPS<b;
}
const int MAX_N=100010;
int N,n;
P ps[MAX_N];
double as[MAX_N];
void solve(){for(int i=0;i<N;i++)scanf("%lf%lf",&ps[i].x,&ps[i].y);vector<P> chs;if(N>1){chs=convex_hull(ps,N);n=chs.size();chs.push_back(chs[0]);}for(int i=0;i<n;i++)as[i]=atan2(chs[i+1]-chs[i]);sort(as,as+n,double_cmp);P p1,p2;while(~scanf("%lf%lf%lf%lf",&p1.x,&p1.y,&p2.x,&p2.y)){if(N<2){puts("GOOD");continue;}int x=upper_bound(as,as+n,atan2(p2-p1),double_cmp)-as;int y=upper_bound(as,as+n,atan2(p1-p2),double_cmp)-as;puts((((p2-p1).det(chs[x]-p1)*(p2-p1).det(chs[y]-p1)>-EPS))?"GOOD":"BAD");}
}
int main(){while(~scanf("%d",&N))solve();return 0;
}

/*题目大意：求出两个凸包之间的最短距离题解：我们先找到一个凸包的上顶点和一个凸包的下定点，以这两个点为起点向下一个点画线，做旋转卡壳，答案一定包含在这个过程中
*/
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <vector>
using namespace std;
double EPS=1e-10;
const double INF=0x3F3F3F3F;
const double PI=acos(-1.0);
double add(double a,double b){if(abs(a+b)<EPS*(abs(a)+abs(b)))return 0;return a+b;
}
struct P{double x,y;P(){}P(double x,double y):x(x),y(y){}P operator + (P p){return P(add(x,p.x),add(y,p.y));}P operator - (P p){return P(add(x,-p.x),add(y,-p.y));}P operator * (double d){return P(x*d,y*d);}double dot(P p){return add(x*p.x,y*p.y);} //点积double det(P p){return add(x*p.y,-y*p.x);}  //叉积
};
bool cmp_y(const P& p,const P& q){if(p.y!=q.y)return p.y<q.y;return p.x<q.x;
}
double dist(P p,P q){return sqrt((p-q).dot(p-q));}
double cross(P a, P b,P c){return(b-a).det(c-a);}
double multi(P a,P b,P c){return(b-a).dot(c-a);}
// 点到线段距离
double point_to_line(P a,P b,P c){if(dist(a,b)<EPS)return dist(b,c);if(multi(a,b,c)<-EPS)return dist(a,c);if(multi(b,a,c)<-EPS)return dist(b,c);return fabs(cross(a,b,c)/dist(a,b));
}
// 线段到线段距离
double line_to_line(P A,P B,P C,P D){double a=point_to_line(A,B,C);double b=point_to_line(A,B,D);double c=point_to_line(C,D,A);double d=point_to_line(C,D,B);return min(min(a,b),min(c,d));
}
void anticlockwise_sort(P* p,int N){for(int i=0;i<N-2;i++){double tmp=cross(p[i],p[i+1],p[i+2]);if(tmp>EPS)return;else if(tmp<-EPS){reverse(p,p+N);return;}}
}
const int MAX_N=10000;
int n,m;
P ps[MAX_N],qs[MAX_N];
void solve(){for(int i=0;i<n;i++)scanf("%lf%lf",&ps[i].x,&ps[i].y);for(int i=0;i<m;i++)scanf("%lf%lf",&qs[i].x,&qs[i].y);anticlockwise_sort(ps,n);anticlockwise_sort(qs,m);int i=0,j=0;for(int k=0;k<n;k++)if(!cmp_y(ps[i],ps[k]))i=k;for(int k=0;k<n;k++)if(cmp_y(qs[j],qs[k]))j=k;double res=INF;ps[n]=ps[0]; qs[m]=qs[0];for(int k=0;k<n;k++){while(cross(ps[i+1],qs[j+1],ps[i])-cross(ps[i+1],qs[j],ps[i])>EPS)j=(j+1)%m;res=min(res,line_to_line(ps[i],ps[i+1],qs[j],qs[j+1]));i=(i+1)%n;}printf("%.5lf\n",res);
}
int main(){while(~scanf("%d%d",&n,&m)&&n+m)solve();return 0;
}

/*题目大意：给出一些点，求出能组成的最大面积的三角形题解：最大三角形一定位于凸包上，因此我们先求凸包，再在凸包上计算，因为三角形在枚举了一条固定边之后，图形面积随着另一个点的位置变换先变大后变小因此我们发现面积递减之后就移动固定边。
*/
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <vector>
using namespace std;
double EPS=1e-10;
const double PI=acos(-1.0);
double add(double a,double b){if(abs(a+b)<EPS*(abs(a)+abs(b)))return 0;return a+b;
}
struct P{double x,y;P(){}P(double x,double y):x(x),y(y){}P operator + (P p){return P(add(x,p.x),add(y,p.y));}P operator - (P p){return P(add(x,-p.x),add(y,-p.y));}P operator * (double d){return P(x*d,y*d);}double dot(P p){return add(x*p.x,y*p.y);} //点积double det(P p){return add(x*p.y,-y*p.x);}  //叉积
};
bool cmp_x(const P& p,const P& q){if(p.x!=q.x)return p.x<q.x;return p.y<q.y;
}
vector<P> convex_hull(P* ps,int n){sort(ps,ps+n,cmp_x);int k=0;vector<P> qs(n*2);for(int i=0;i<n;i++){while((k>1)&&(qs[k-1]-qs[k-2]).det(ps[i]-qs[k-1])<=0)k--;qs[k++]=ps[i];}for(int i=n-2,t=k;i>=0;i--){while(k>t&&(qs[k-1]-qs[k-2]).det(ps[i]-qs[k-1])<=0)k--;qs[k++]=ps[i];}qs.resize(k-1);return qs;
}
double cross(P A,P B,P C){return(B-A).det(C-A);}
const int MAX_N=50010;
int N;
P ps[MAX_N];
void solve(){for(int i=0;i<N;i++)scanf("%lf%lf",&ps[i].x,&ps[i].y);vector<P> qs=convex_hull(ps,N);N=qs.size(); int ans=0;for(int i=1;i<(N+1)/2;i++){ //枚举跨度 int p1=(i+1)%N;for(int p3=0;p3<N;p3++){int p2=(p3+i)%N;int prev=abs(cross(qs[p3],qs[p2],qs[p1]));for(++p1;p1!=p2&&p1!=p3;++p1){if(p1==N)p1=0;int cur=abs(cross(qs[p3],qs[p2],qs[p1]));ans=max(ans,prev);if(cur<=prev)break;prev=cur;}--p1;if(p1==-1)p1+=N;}}printf("%d.%s\n",ans/2,ans%2==1?"50":"00");
}
int main(){while(~scanf("%d",&N)&&N>0)solve();return 0;
}

/*题目大意:给出一些点，请删去一个点，使得包围这些点用的线长最短题解：去掉的点肯定是凸包上的点，所以枚举凸包上的点去掉，再计算面积即可。
*/
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <vector>
#include <cstring>
using namespace std;
struct P{int x,y;int id;P(){}P(double x,double y):x(x),y(y){}P operator + (P p){return P(x+p.x,y+p.y);}P operator - (P p){return P(x-p.x,y-p.y);}P operator * (double d){return P(x*d,y*d);}int dot(P p){return x*p.x+y*p.y;} //点积int det(P p){return x*p.y-y*p.x;}  //叉积
};
bool cmp_x(const P& p,const P& q){if(p.x!=q.x)return p.x<q.x;return p.y<q.y;
}
vector<P> convex_hull(P* ps,int n){sort(ps,ps+n,cmp_x);int k=0;vector<P> qs(n*2);for(int i=0;i<n;i++){while((k>1)&&(qs[k-1]-qs[k-2]).det(ps[i]-qs[k-1])<=0)k--;qs[k++]=ps[i];}for(int i=n-2,t=k;i>=0;i--){while(k>t&&(qs[k-1]-qs[k-2]).det(ps[i]-qs[k-1])<=0)k--;qs[k++]=ps[i];}qs.resize(k-1);return qs;
}
int cross(P a, P b,P c){return(b-a).det(c-a);}
int compute_area(P A,P B,P C){int res=cross(A,B,C);if(res<0){return -res;}return res;
}
int compute_area(const vector<P>& ps){int total=0;for(int i=2;i<ps.size();i++){total+=compute_area(ps[0],ps[i-1],ps[i]);}return total;
}
const int MAX_N=100010;
int N;
P p[MAX_N],q[MAX_N];
void solve(){for(int i=0;i<N;i++){scanf("%d%d",&p[i].x,&p[i].y);p[i].id=i;}memcpy(q,p,N*sizeof(P));vector<P> ps=convex_hull(p,N);int ans=0x3f3f3f3f;for(int i=0;i<ps.size();i++){memcpy(p,q,N*sizeof(P));swap(p[ps[i].id],p[N-1]);ans=min(ans,compute_area(convex_hull(p,N-1)));}printf("%d.%s\n",ans/2,ans%2==1?"50":"00");
}
int main(){while(~scanf("%d",&N),N)solve();return 0;
}