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 rep(i, a, b) for(int i = (a); i <= (b); ++i)
#define per(i, a, b) for(int i = (a); i >= (b); --i)
#define debug(x) cerr << #x << ' ' << x << endl;
using namespace std;

typedef long long ll;
const int mod = 1e9+7;
const int MAXN = 2e5 + 7;
const double EPS=1e-8;
inline int sign(double a){return a<-EPS?-1:a>EPS;}
inline int cmp(double a,double b){return sign(a-b);}
//点
struct P{
    double x,y;
    P(){}
    P(double _x,double _y):x(_x),y(_y){}
    P operator + (P p){return P(x+p.x,y+p.y);}
    P operator - (P p){return P(x-p.x,y-p.y);}
    P operator * (double k){return P(x*k,y*k);}
    P operator / (double k){return P(x/k,y/k);}
    double dot(P p){return x*p.x+y*p.y;}
    double det(P p){return x*p.y-y*p.x;}
    double distTo(P p){return (*this-p).abs();}
    double alpha(){return atan2(y,x);}
    void read(){scanf("%lf%lf", &x, &y);}
    void write(){printf("%.0lf %.0lf\n", x, y);}
    double abs(){return sqrt(abs2());}
    double abs2(){return x*x+y*y;}
    P rot90(){return P(-y,x);}
    P unit(){return *this/abs();}
    int quad(){return sign(y)==1||(sign(y)==0&&sign(x)>=0);}
    P rot(double an){return P(x*cos(an)-y*sin(an),x*sin(an)+y*cos(an));}
    bool operator < (P p)const{int c=cmp(x,p.x);if(c)return c==-1;return cmp(y,p.y)==-1;}
    bool operator == (P o)const{return cmp(x,o.x)==0&&cmp(y,o.y)==0;}
}s[MAXN];

int main(int argc, char const *argv[])
{
    int n;
    scanf("%d", &n);
    rep(i, 0, n-1)    s[i].read();
    int j = 2;
    double ans = 0;
    rep(i, 0, n-1) {
        while((s[i]-s[j]).det(s[i+1]-s[j]) <= (s[i]-s[(j+1)%n]).det(s[i+1]-s[(j+1)%n]))  j = (j+1)%n;
        ans = max(ans, max((s[i]-s[j]).abs(), (s[i+1]-s[j]).abs()));
    }
    printf("%.16lf\n", ans);
    return 0;
}