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>
using namespace std;

typedef long long ll;
#define REP(i,n) for(int i=0,_n=(int)(n);i<_n;++i)
#define ALL(v) (v).begin(),(v).end()
#define CLR(t,v) memset(t,(v),sizeof(t))
template<class T1,class T2>ostream& operator<<(ostream& os,const pair<T1,T2>&a){return os<<"("<<a.first<<","<<a.second<< ")";}
template<class T>void pv(T a,T b){for(T i=a;i!=b;++i)cout<<(*i)<<" ";cout<<endl;}
template<class T>void chmin(T&a,const T&b){if(a>b)a=b;}
template<class T>void chmax(T&a,const T&b){if(a<b)a=b;}


typedef long double D;
const D PI = acos(-1.0);
const D EPS = 1e-10;
class P { public:
  D x, y;
  P(D x=0, D y=0) : x(x), y(y) {}

  P& operator+=(const P& o) { x += o.x; y += o.y; return *this; }
  P& operator-=(const P& o) { x -= o.x; y -= o.y; return *this; }
  P& operator*=(const P& o) { return *this = {x*o.x - y*o.y, x*o.y + y*o.x}; }
  P& operator*=(const D& r) { x *= r; y *= r; return *this; }
  P& operator/=(const D& r) { x /= r; y /= r; return *this; }
  P operator-() const { return {-x, -y}; }

  D norm() const { return x*x + y*y; }
  D abs() const { return sqrt(norm()); }
  D arg() const { return atan2(y, x); }
  bool isZero() const { return std::abs(x) < EPS && std::abs(y) < EPS; }
  /** 象限 */
  int orth() const { return y >= 0 ? (x >= 0 ? 1 : 2) : (x < 0 ? 3 : 4); }
  static P polar(const D& rho, const D& theta = 0) { return {rho * cos(theta), rho * sin(theta)}; }
};
std::ostream &operator<<(std::ostream &os, P const &p) { return os << "(" << p.x << ", " << p.y << ")"; }
std::istream &operator>>(std::istream &is, P &p) { D a, b; is >> a >> b; p = P(a, b); return is; }
P operator+(const P& p, const P& q) { return P(p) += q; }
P operator-(const P& p, const P& q) { return P(p) -= q; }
P operator*(const P& p, const P& q) { return P(p) *= q; }
P operator*(const P& p, const D& r) { return P(p) *= r; }
P operator/(const P& p, const D& r) { return P(p) /= r; }
P operator*(const D& r, const P& p) { return P(p) *= r; }
P operator/(const D& r, const P& p) { return P(p) /= r; }
D crs(const P& a, const P& b){ return a.x*b.y - a.y*b.x; }
D dot(const P& a, const P& b){ return a.x*b.x + a.y*b.y; }
int signum(D x) {return x > EPS ? +1 : x < -EPS ? -1 : 0;}


// 凸包の直径
D convexDiamiter(const vector<P>& ps){
  int N = (int)ps.size();
  int i = 0, j = 0;
  for (int k = 1; k < N; k++) {
    if (ps[k].y > ps[i].y) i = k;
    if (ps[k].y < ps[j].y) j = k;
  }
  int si = i, sj = j;
  D max_dist = (ps[i] - ps[j]).norm();
  do {
    if (crs(ps[(i+1)%N] - ps[i], ps[(j+1)%N] - ps[j]) >= 0) j = (j+1) % N;
    else i = (i+1) % N;
    if ((ps[i] - ps[j]).norm() > max_dist) {
      max_dist = (ps[i] - ps[j]).norm();
    }
  } while( si != i || sj != j );
  return sqrt(max_dist);
}


int main2() {
  int N; cin >> N;
  vector<P> ps(N);
  REP(i, N) cin >> ps[i];
  double ans = (double)convexDiamiter(ps);
  printf("%.10f\n", ans);
  return 0;
}

int main() {

#ifdef LOCAL
  for (;!cin.eof();cin>>ws)
#endif
    main2();
  return 0;
}