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<string>
#include<algorithm>
#include<map>
#include<set>
#include<utility>
#include<vector>
#include<cmath>
#include<cstdio>
#include<complex>
#define loop(i,a,b) for(int i=a;i<b;i++) 
#define rep(i,a) loop(i,0,a)
#define pb push_back
#define mp make_pair
#define it ::iterator
#define all(in) in.begin(),in.end()
const double PI=acos(-1);
const double EPS=1e-10;
using namespace std;
typedef complex<double> P;
typedef vector<P> G;
struct L : public vector<P> {
  L(const P &a, const P &b) {
    push_back(a); push_back(b);
  }
};
#define curr(P, i) P[i]
#define next(P, i) P[(i+1)%P.size()]
#define diff(P, i) (next(P, i) - curr(P, i))
namespace std {
	bool operator < (const P& a, const P& b) {
		//return real(a) != real(b) ? real(a) < real(b) : imag(a) < imag(b);
		return imag(a) != imag(b) ? imag(a) < imag(b) : real(a) < real(b);  
	}
}
P pin(){
	double x,y;
	cin>>x>>y;
	P p(x,y);
	return p;
}
void PIN(P* a,int n){
	rep(i,n)a[i]=pin();
}
double dot(P a,P b){
	return real(conj(a)*b);
}
double cross(P a,P b){
	return imag(conj(a)*b);
}
int ccw(P a, P b, P c) {
  b -= a; c -= a;
  if (cross(b, c) > 0)   return +1;       // counter clockwise
  if (cross(b, c) < 0)   return -1;       // clockwise
  if (dot(b, c) < 0)     return +2;       // c--a--b on line
  if (norm(b) < norm(c)) return -2;       // a--b--c on line
  return 0;
}
P projection(L a,P p){
	double t=dot(p-a[0],a[0]-a[1])/norm(a[0]-a[1]);
	return a[0]+t*(a[0]-a[1]);
}
P reflection(L a,P p){
	return p+2.0*(projection(a,p)-p);
}
bool intersectLL(const L &l, const L &m) {
  return abs(cross(l[1]-l[0], m[1]-m[0])) > EPS || // non-parallel
         abs(cross(l[1]-l[0], m[0]-l[0])) < EPS;   // same line
}
bool intersectLS(const L &l, const L &s) {
  return cross(l[1]-l[0], s[0]-l[0])*       // s[0] is left of l
         cross(l[1]-l[0], s[1]-l[0]) < EPS; // s[1] is right of l
}
bool intersectLP(const L &l, const P &p) {
  return abs(cross(l[1]-p, l[0]-p)) < EPS;
}
bool intersectSS(const L &s, const L &t) {
  return ccw(s[0],s[1],t[0])*ccw(s[0],s[1],t[1]) <= 0 &&
         ccw(t[0],t[1],s[0])*ccw(t[0],t[1],s[1]) <= 0;
}
bool intersectSP(const L &s, const P &p) {
  return abs(s[0]-p)+abs(s[1]-p)-abs(s[1]-s[0]) < EPS; // triangle inequality
}
P intersect_ls(L a,L b){
	double t1=abs(cross(a[1]-a[0],b[0]-a[0]));
	double t2=abs(cross(a[1]-a[0],b[1]-a[0]));
	return b[0]+(b[1]-b[0])*t1/(t1+t2);
}
double distanceLP(const L &l, const P &p) {
  return abs(p - projection(l, p));
}
double distanceLL(const L &l, const L &m) {
  return intersectLL(l, m) ? 0 : distanceLP(l, m[0]);
}
double distanceLS(const L &l, const L &s) {
  if (intersectLS(l, s)) return 0;
  return min(distanceLP(l, s[0]), distanceLP(l, s[1]));
}
double distanceSP(const L &s, const P &p) {
  const P r = projection(s, p);
  if (intersectSP(s, r)) return abs(r - p);
  return min(abs(s[0] - p), abs(s[1] - p));
}
double distanceSS(const L &s, const L &t) {
  if (intersectSS(s, t)) return 0;
  return min(min(distanceSP(s, t[0]), distanceSP(s, t[1])),
             min(distanceSP(t, s[0]), distanceSP(t, s[1])));
}
double area(const G &g){
    double S =0;
    for(int i =0;i <g.size();i++){
        S +=(cross(g[i],g[(i+1)%g.size()]));
    }
    return abs(S/2.0);
}
bool inconvex(const G &g){
	int n=g.size();
	rep(i,n)if(ccw(g[(i+n-1)%n],g[i%n],g[(i+1)%n])==-1)return false;
	return true;
}
int contain(const G& g, const P& p) {
	bool in = false;
	int n=g.size();
	rep(i,n){
		P a=g[i%n]-p;
		P b=g[(i+1)%n]-p;
		if(imag(a)>imag(b))swap(a, b);
		if(imag(a)<=0&&0<imag(b))if(cross(a,b)<0)in=!in;
		if(cross(a,b)==0&&dot(a,b)<=0)return 1;//ON
	}
	return in?2:0;//IN : OUT;
}
G convex_hull(G &ps) {
    int n=ps.size(),k=0;
	sort(ps.begin(), ps.end());
	G ch(2*n);
	for(int i=0;i<n;ch[k++]=ps[i++])//lower-hull
		while(k>=2&&ccw(ch[k-2],ch[k-1],ps[i])==-1)--k;//<=0  ->  ==-1
	for(int i=n-2,t=k+1;i>=0;ch[k++]=ps[i--])//upper-hull
		while(k>=t&&ccw(ch[k-2],ch[k-1],ps[i])==-1)--k;//
	ch.resize(k-1);
	return ch;
}
double convex_diameter(const G &pt) {
  const int n = pt.size();
  int is = 0, js = 0;
  for (int i = 1; i < n; ++i) {
    if (imag(pt[i]) > imag(pt[is])) is = i;
    if (imag(pt[i]) < imag(pt[js])) js = i;
  }
  double maxd = norm(pt[is]-pt[js]);

  int i, maxi, j, maxj;
  i = maxi = is;
  j = maxj = js;
  do {
    if (cross(diff(pt,i), diff(pt,j)) >= 0) j = (j+1) % n;
    else i = (i+1) % n;
    if (norm(pt[i]-pt[j]) > maxd) {
      maxd = norm(pt[i]-pt[j]);
      maxi = i; maxj = j;
    }
  } while (i != is || j != js);
  return sqrt(maxd); /* farthest pair is (maxi, maxj). */
}//convex_diameter(g)
  int main(){
	int n;
	cin>>n;
	G g(n);
	PIN(&g[0],n);
	printf("%.9f\n",convex_diameter(g));
}