#pragma GCC optimize("O2")
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include<cstring>
#include<string>
#include<cmath>
#include<queue>
#define cls(x) memset(x,0,sizeof(x))
using namespace std;
template<typename tp>void read(tp & dig)
{
char c=getchar();dig=0;
while(!isdigit(c))c=getchar();
while(isdigit(c))dig=dig*10+c-'0',c=getchar();
}
//begin
const double eps=1e-9,pi=4*atan(1);
struct Point
{
double x,y;
int id;
Point(){}
Point(double _x,double _y){x=_x,y=_y;}
};
struct Line
{
Point u,v;
Line(){}
Line(Point _u,Point _v){u=_u,v=_v;}
};
struct Circle
{
Point c;double r;
Circle(){}
Circle(Point _c,double _r){c=_c;r=_r;}
Point point (double a){return Point(c.x+cos(a)*r,c.y+sin(a)*r);}
};
typedef Point Vector ;
Vector operator - (Point a,Point b){return Point(a.x-b.x,a.y-b.y);}
bool operator < (const Point &a,const Point b){return a.x<b.x||(a.x==b.x&&a.y<a.y);}
int dcmp(double x){if(fabs(x)<eps) return 0;else return x>0?1:-1;}
double Cross(Vector a,Vector b){return a.x*b.y-a.y*b.x;}
double Dot(Vector a,Vector b){return a.x*b.x+a.y*b.y;}
double xmul(Vector p0,Vector p1,Vector p2){return Cross(p1-p0,p2-p0);}
bool cmp_point(Point a,Point b){return a.x<b.x||(a.x==b.x&&a.y<b.y);}
bool cross_seg(int n,Point *s,Point *t,Point a,Point b)//直线穿过线段
{
if((!dcmp(a.x-b.x))&&(!dcmp(a.y-b.y))) return 0;
for(int i=1;i<=n;i++)
if(xmul(s[i],a,b)*xmul(t[i],a,b)>eps) return 0;
return 1;
}
bool segcross(Point a1,Point a2,Point b1,Point b2)
{
double t1=xmul(a1,a2,b1),t2=xmul(a1,a2,b2),
t3=xmul(b1,b2,a1),t4=xmul(b1,b2,a2);
return dcmp(t1)*dcmp(t2)<0&&dcmp(t3)*dcmp(t4)<0;
}
bool onseg(Point p,Point a1,Point a2)
{
return dcmp(Cross(a1-p,a2-p))==0&&dcmp(Dot(a1-p,a2-p))<0;
//前者因为sin180=0 后者点积为负cos为负>90°然后就可以判断出在两个端点组成的长方形之内。
}
double Area(Point *p,int n)//多边形的有向面积 不一定为凸
{
double area=0;
for(int i=2;i<=n-1;i++)
area+=Cross(p[i]-p[1],p[i+1]-p[1]);
return area/2;
}
bool sameseg(Point p,Point a1,Point a2){return dcmp(Cross(a1-p,a2-p))==0;}
Point line_intersection(Point p1,Point p2,Point p3,Point p4)
{
double a1=p1.y-p2.y,b1=p2.x-p1.x,c1=Cross(p1,p2);
double a2=p3.y-p4.y,b2=p4.x-p3.x,c2=Cross(p3,p4);
return Point((c1*b2-c2*b1)/(a2*b1-a1*b2),(a2*c1-a1*c2)/(a1*b2-a2*b1));
}
bool LineParallel(Point p1,Point p2,Point p3,Point p4){return dcmp(Cross(Point(p2-p1),Point(p4-p3)))==0;}
double dist(Point a,Point b){return sqrt(Dot(a-b,a-b));}
bool segconcross(Line a,Line b)
{
return segcross(a.u,a.v,b.u,b.v)||
onseg(a.u,b.u,b.v)||onseg(a.v,b.u,b.v)
||onseg(b.u,a.u,a.v)||onseg(b.v,a.u,a.v);
}
bool inrectangle(Point p1,Point p2,Point p3,Point p4,Line ls)
{
int ym=max(p1.y,p3.y),ys=min(p1.y,p3.y);
int xm=max(p1.x,p3.x),xs=min(p1.x,p3.x);
int lym=max(ls.u.y,ls.v.y),lys=min(ls.u.y,ls.v.y);
int lxm=max(ls.u.x,ls.v.x),lxs=min(ls.u.x,ls.v.x);
return (xm>=lxm&&lxs>=xs&&ym>=lym&&lys>=ys);
}
//如果不允许点在凸包的边上的话把两个<=变成<
int ConvexHull(Point *p,int n,Point *ch)
{
sort(p+1,p+n+1,cmp_point);
int cnt=0;
for(int i=1;i<=n;i++)
{
while(cnt>1&&Cross(ch[cnt]-ch[cnt-1],p[i]-ch[cnt-1])<=0) cnt--;//注意这里cnt>1 保证栈有两个元素
ch[++cnt]=p[i];
}
int k=cnt;
for(int i=n-1;i>=1;i--)
{
while(cnt>k&&Cross(ch[cnt]-ch[cnt-1],p[i]-ch[cnt-1])<=0) cnt--;
ch[++cnt]=p[i];
}
return cnt;
}
//end
int main()
{
return 0;
}