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 <bits/stdc++.h>
#define syosu(x) fixed<<setprecision(x)
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> P;
typedef pair<double,double> pdd;
typedef pair<ll,ll> pll;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<double> vd;
typedef vector<vd> vvd;
typedef vector<ll> vl;
typedef vector<vl> vvl;
typedef vector<string> vs;
typedef vector<P> vp;
typedef vector<vp> vvp;
typedef vector<pll> vpll;
typedef pair<int,P> pip;
typedef vector<pip> vip;
const int inf=1<<30;
const ll INF=1ll<<60;
const double pi=acos(-1);
const double eps=1e-9;
const ll mod=1e9+7;
const int dx[4]={0,1,0,-1},dy[4]={1,0,-1,0};

typedef complex<double> C;
typedef pair<C,C> pp;
typedef vector<C> VP;
typedef vector<pp> VPP;
#define eq(a,b) (fabs(a-b)<eps)
#define veq(a,b) (eq(a.real(),b.real())&&eq(a.imag(),b.imag()))
const C O{0,0};

void In(C& p){
	double x,y;
	cin>>x>>y;
	p=C(x,y);
}

void Out(C p){
	cout<<(int)p.real()<<' '<<(int)p.imag()<<endl;
}

double Dot(C p,C q){
	return p.real()*q.real()+p.imag()*q.imag();
}

double Det(C p,C q){
	return p.real()*q.imag()-q.real()*p.imag();
}

double Diameter(VP p){
	int n=p.size(),I=0,J=0;
	for(int i=1;i<n;i++){
		if(p[i].imag()>p[I].imag()) I=i;
		if(p[i].imag()<p[J].imag()) J=i;
	}
	double res=abs(p[I]-p[J]);
	int i=I,j=J;
	do{
		if(Det(p[(i+1)%n]-p[i],p[(j+1)%n]-p[j])>0) j=(j+1)%n;
		else i=(i+1)%n;
		if(abs(p[i]-p[j])>res) res=abs(p[i]-p[j]);
	}while(i!=I||j!=J);
	return res;
}

int n,q;
VP a;
C x;

int main(){
	cin>>n;
	a=VP(n);
	for(auto &i:a) In(i);
	cout<<syosu(9)<<Diameter(a)<<endl;
}