Problem: Find the diameter of a convex polygon g. In other words, find a pair of points that have maximum distance between them.

Constraints

* 3 ≤ n ≤ 80000
* -100 ≤ xi, yi ≤ 100
* No point in the g will occur more than once.

Input


n
x1 y1
x2 y2
:
xn yn


The first integer n is the number of points in g.

In the following lines, the coordinate of the i-th point pi is given by two real numbers xi and yi. The coordinates of points are given in the order of counter-clockwise visit of them. Each value is a real number with at most 6 digits after the decimal point.

Output

Print the diameter of g in a line. The output values should be in a decimal fraction with an error less than 0.000001.

Examples

Input

3
0.0 0.0
4.0 0.0
2.0 2.0


Output

4.00


Input

4
0.0 0.0
1.0 0.0
1.0 1.0
0.0 1.0


Output

1.414213562373

Solution (in CPP):

Solution:
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<vector>
#include<queue>
#include<algorithm>
#define N 100010
#define inf 1000000000
#define eps 1e-10
#define double long double
using namespace std;
struct point{
	double x,y;
	void read(){scanf("%Lf%Lf",&x,&y);}
	void write(){printf("%.10Lf %.10Lf\n",x,y);}
	point(double X=0.0,double Y=0.0):x(X),y(Y){}
	bool operator <(const point a)const{return y==a.y?x<a.x:y<a.y;}
	point operator +(const point a)const{return point(x+a.x,y+a.y);}
	point operator -(const point a)const{return point(x-a.x,y-a.y);}
	point operator *(double a)const{return point(x*a,y*a);}
	point operator /(double a)const{return point(x/a,y/a);}
	bool operator ==(const point a)const{return abs(x-a.x)<=eps && abs(y-a.y)<=eps;}
};
double dis(point a,point b=point(0,0)){return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));}
double dot(point a,point b){return a.x*b.x+a.y*b.y;}
double cross(point a,point b){return a.x*b.y-a.y*b.x;}
double cross3(point o,point a,point b){return cross(a-o,b-o);}
bool jiao(point a,point b,point c,point d)
{
	if(min(a.x,b.x)>max(c.x,d.x) || min(a.y,b.y)>max(c.y,d.y)) return false;
	if(min(c.x,d.x)>max(a.x,b.x) || min(c.y,d.y)>max(a.y,b.y)) return false;
	if(cross3(a,b,c)*cross3(a,b,d)>0) return false;
	if(cross3(c,d,a)*cross3(c,d,b)>0) return false;
	return true;
}
short int on_line(point o,point a,point b)//oa->ob 0:D 1:U 2:O
{
	if(dot(a-o,b-o)<-eps) return 0;
	return 1+(dis(a-o)>=dis(b-o));
}
void make_rand(double &x){x=x+(1ll*rand()%10+1)/10.0*eps;}
struct Line{
	point a,b;
	Line(point x=point(0,0),point y=point(0,0)):a(x),b(y){}
	void read(){a.read();b.read();}
	bool operator &&(const Line u){return jiao(a,b,u.a,u.b);}
	point operator &(const Line u)
	{
		if(a==u.a || a==u.b) return a;
		if(b==u.a || b==u.b) return b;
		double x1=a.x,x2=b.x,x3=u.a.x,x4=u.b.x;
		double y1=a.y,y2=b.y,y3=u.a.y,y4=u.b.y;
		if(x3==x4) return point(x3,(y2-y1)*x3/(x2-x1)+(y1*x2-x1*y2)/(x2-x1));
		if(x1==x2) return point(x1,(y4-y3)*x1/(x4-x3)+(y3*x4-x3*y4)/(x4-x3));
		make_rand(x2);
		double x=((y3*x4-y4*x3)*(x2-x1)-(y1*x2-y2*x1)*(x4-x3))/((y2-y1)*(x4-x3)-(y4-y3)*(x2-x1));
		double y=(y2-y1)*x/(x2-x1)+(y1*x2-x1*y2)/(x2-x1);
		return point(x,y);
	}
};
point project(Line a,point p)
{
	point &p1=a.a,&p2=a.b;
	double len=dot(p2-p1,p-p1)/dis(p2-p1);
	return p1+(p2-p1)/dis(p2-p1)*len;
}
double dis(Line a,Line b)
{
	if(a && b) return 0;
	double ans=min(min(dis(a.a,b.a),dis(a.a,b.b)),min(dis(a.b,b.a),dis(a.b,b.b)));
	point res=project(a,b.a);
	if(on_line(a.a,a.b,res)==2) ans=min(ans,dis(res,b.a));
	res=project(a,b.b);
	if(on_line(a.a,a.b,res)==2) ans=min(ans,dis(res,b.b));
	res=project(b,a.a);
	if(on_line(b.a,b.b,res)==2) ans=min(ans,dis(res,a.a));
	res=project(b,a.b);
	if(on_line(b.a,b.b,res)==2) ans=min(ans,dis(res,a.b));
	return ans;
}
int sta[N],tot;
struct polygon{
	vector<point>p;
	void read(int n=-1)
	{
		if(n<0) scanf("%d",&n);
		for(int i=1;i<=n;i++)
		{
			point a;
			a.read();
			p.push_back(a);
		}
	}
	double area(void)
	{
		double ans=0;
		for(int i=1;i<p.size();i++)
		ans+=cross3(p[0],p[i-1],p[i])/2;
		return abs(ans);
	}
	bool convex(void)
	{
		short int right=-1;
		for(int i=2;i<p.size();i++)
		if(right<0) right=cross3(p[i-2],p[i-1],p[i])>=0;
		else if(right!=(cross3(p[i-2],p[i-1],p[i])>=0)) return false;
		if(right!=(cross3(p[p.size()-2],p[p.size()-1],p[0])>=0)) return false;
		return true;
	}
	short int in_polygon(const point x)
	{
		Line l(x,x+point(inf,(double)inf*1.414213562373));
		make_rand(l.b.y);
		int cnt=0;
		for(int i=1;i<p.size();i++)
		{
			Line u(p[i-1],p[i]);
			if(u && l)
			{
				cnt++;
				if(abs(cross3(p[i-1],p[i],x))<eps && on_line(p[i-1],p[i],x)==2) return 1;
			}
		}
		Line u(p[p.size()-1],p[0]);
		if(u && l)
		{
			cnt++;
			if(abs(cross3(p[p.size()-1],p[0],x))<eps && on_line(p[p.size()-1],p[0],x)==2) return 1;
		}
		return cnt%2==0?0:2; 
	}
	void make_convex(void)
	{
		sort(p.begin(),p.end());
		vector<point>s;
		tot=0;
		for(int i=0;i<p.size();i++)
		{
			while(tot>=2 && cross3(p[i],p[sta[tot-1]],p[sta[tot-2]])>=eps) tot--;
			sta[tot++]=i;
		}
		int swp=tot;
		for(int i=p.size()-2;i>=0;i--)
		{
			while(tot-swp>=1 && cross3(p[i],p[sta[tot-1]],p[sta[tot-2]])>=eps) tot--;
			sta[tot++]=i;
		}
		for(int i=0;i<tot-1;i++)
		s.push_back(p[sta[i]]);
		sort(s.begin(),s.end());
		p=s; 
	}
	double diameter(void)
	{
		int l=0,r=1;
		double ans=dis(p[0],p[1]);
		do
		{
			while(dis(p[l],p[r])<=dis(p[l],p[(r+1)%p.size()]))
			r=(r+1)%p.size(),ans=max(ans,dis(p[l],p[r]));
			l=(l+1)%p.size();
			ans=max(ans,dis(p[l],p[r]));
		}
		while(l!=0);
//		for(int l=0;l<p.size();l++)
//			for(int r=l+1;r<p.size();r++)
//			if(dis(p[l],p[r])>ans) printf("%d %d\n",l,r);
		return ans;
	}
}p;
int main()
{
	int n;
	scanf("%d",&n);
	
	srand(n);
	p.read(n);
//	p.make_convex();
	printf("%.10Lf\n",p.diameter());
	return 0;
}