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<complex>
#include<vector>
#include<iomanip>
#include<algorithm>

using namespace std;

typedef complex<double> P;
typedef vector<P> G;
#define EPS 1e-9
#define shosu(x) fixed<<setprecision(x)
#define diff(P,i) (P[(i+1)%P.size()] - P[i])

struct L : public vector<P> {
  L(const P &a, const P &b) {
    push_back(a); push_back(b);
  }
  L(){;}
};
struct C {
  P c;double r;
  C(const P &c,double r):c(c),r(r){}
};

namespace std{//演算子の定義
  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)
  }
  bool operator == (const P& a,const P& b) {
    return a.real()==b.real() && a.imag()==b.imag();
  }
}

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) { //3点の関係性
  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;                               // a--c--b on line
}

P projection(const L &l, const P &p) { //pの直線l上の射影の点
  double t=dot(p-l[0],l[0]-l[1])/norm(l[0]-l[1]);
  return l[0]+t*(l[0]-l[1]);
}

P reflection(const L &l, const P &p) { //点pの直線lに関して対称な点
  return p+2.0*(projection(l,p)-p);
}

bool isorthogonal(const L &l, const L &m) { //2直線の直交判定
  return fabs(dot(l[1]-l[0], m[1]-m[0])) < EPS;
}

bool isparallel(const L &l, const L &m) { //2直線の平行判定
  return fabs(cross(l[1]-l[0],m[1]-m[0])) < EPS;
}

bool intersectLL(const L &l, const L &m) { //2直線の交差判定
  return !isparallel(l,m);
}

bool intersectSS(const L &s, const L &t) { //2線分の交差判定(完全に交差してないとだめ)
  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) { //直線と点の交差判定
  return abs(s[0]-p)+abs(s[1]-p)-abs(s[1]-s[0]) < EPS; // triangle inequality
}

P crosspointSS(const L &a, const L &b) { //2線分の交点
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 distanceSP(const L &s, const P &p) {//直線と点との距離
  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) {//2線分の距離
  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) { //多角形の面積
    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 isconvex(const G &g) { //凸かどうか(全ての内角の大きさが180度以下)
	int n=g.size();
  for(int i = 0; i < n; i++)
    if(ccw(g[(i+n-1)%n],g[i%n],g[(i+1)%n])==-1)return false;
	return true;
}

int inconvex(const G& g, const P& p) { //多角形と点の関係
	bool in = false;
	int n = g.size();
	for(int i = 0; i < n; i++){
		P a=g[i%n]-p;
		P b=g[(i+1)%n]-p;
		if(imag(a)>imag(b))swap(a, b);
		if(imag(a)<EPS&&0<imag(b))if(cross(a,b)<0)in=!in;
		if(abs(cross(a,b))<EPS&&dot(a,b)<EPS)return 1;//ON
	}
	return in?2:0;//IN : OUT;
}

G convex_hull(G &ps) { // 凸包(点集合 P の全ての点を含む最小の凸多角形)
  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) { //凸多角形の直径(最遠頂点対間距離)
  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)


P inP(){
  double x,y;
  cin >> x >> y;
  P p(x,y);
  return p;
}

L inL(){
  P p1=inP();
  P p2=inP();
  L l(p1,p2);
  return l;
}

int main(){

  int n;
  cin >> n;
  G g;
  for(int i = 0; i< n; i++){
    P p = inP();
    g.push_back(p);
  }

  cout << shosu(10) << convex_diameter(g) << endl;
}