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 _overload(_1,_2,_3,name,...) name
#define _rep(i,n) _range(i,0,n)
#define _range(i,a,b) for(int i=int(a);i<int(b);++i)
#define rep(...) _overload(__VA_ARGS__,_range,_rep,)(__VA_ARGS__)

#define _rrep(i,n) _rrange(i,n,0)
#define _rrange(i,a,b) for(int i=int(a)-1;i>=int(b);--i)
#define rrep(...) _overload(__VA_ARGS__,_rrange,_rrep,)(__VA_ARGS__)

#define _all(arg) begin(arg),end(arg)
#define uniq(arg) sort(_all(arg)),(arg).erase(unique(_all(arg)),end(arg))
#define getidx(ary,key) lower_bound(_all(ary),key)-begin(ary)
#define clr(a,b) memset((a),(b),sizeof(a))
#define bit(n) (1LL<<(n))
#define popcount(n) (__builtin_popcountll(n))

using namespace std;

template<class T>bool chmax(T &a, const T &b) { return (a<b)?(a=b,1):0;}
template<class T>bool chmin(T &a, const T &b) { return (b<a)?(a=b,1):0;}

using ll=long long;
using R=long double;
const R EPS=1e-9L; // [-1000,1000]->EPS=1e-8 [-10000,10000]->EPS=1e-7
inline int sgn(const R& r){return(r > EPS)-(r < -EPS);}
inline R sq(R x){return sqrt(max(x,0.0L));}

const int dx[8]={1,0,-1,0,1,-1,-1,1};
const int dy[8]={0,1,0,-1,1,1,-1,-1};

// Description: ????????????
// Verifyed: various problem 
const R INF = 1E40L;
const R PI = acos(-1.0L);
using P=complex<R>;
const P O=0.0L;
using L=struct{P s,t;};
using VP=vector<P>;
using C=struct{P p;R c;};

istream& operator >> (istream& is,P& p){ R x,y;is >> x >> y; p=P(x,y); return is;}
ostream& operator << (ostream& os,P& p){ os << real(p) << " " << imag(p); return os;}

inline R dot(P o,P a,P b){return real(conj(a-o)*(b-o));}
inline R det(P o,P a,P b){return imag(conj(a-o)*(b-o));}
inline int sdot(P o,P a,P b){return sgn(dot(o,a,b));}
inline int sdet(P o,P a,P b){return sgn(det(o,a,b));}

//projection verify AOJ CGL_1_A
P proj(L l,P p){ R u=real((p-l.s)/(l.t-l.s)); return (1-u)*l.s+u*l.t;}

// farthest point pair Verify AOJ CGL_4_B
R fpp(VP pol){
	int n=pol.size(),i=0,j=0;
	if(n<=2) return abs(pol[0]-pol[1]);
	R res=0.0;
	auto cmp_x=[](P a,P b)->bool{return sgn(real(a)-real(b))?sgn(real(a)-real(b))<0:sgn(imag(a)-imag(b)<0);};
	
	rep(k,n){
		if(!cmp_x(pol[i],pol[k]))i=k;
		if(cmp_x(pol[j],pol[k]))j=k;
	}

	int si=i,sj=j;
	while(i!=sj||j!=si){
		chmax(res,abs(pol[i]-pol[j]));
		if(sdet(O,pol[(i+1)%n]-pol[i],pol[(j+1)%n]-pol[j])<0)
			i=(i+1)%n;
		else 
			j=(j+1)%n;
	}
	return res;
}

int main(void){
	int n;
	cin >> n;
	VP pol;
	rep(i,n){
		P p;
		cin >> p;
		pol.push_back(p);
	}
	cout.precision(10);
	cout << fixed << fpp(pol) << endl;
	return 0;
}