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>
#include<iostream>
#include<stdio.h>
#include<cmath>
#include<vector>
#include<algorithm>
using namespace std;
typedef double db;
const db eps = 1e-15, pi = acos(-1);
int sign(db x) {return x < -eps ? -1 : x > eps;}
int cmp(db x, db y) {return sign(x - y);}
int intersect(db l1, db r1, db l2, db r2) {
    if (l1 > r1) swap(l1, r1); if (l2 > r2) swap(l2, r2); return cmp(r1, l2) != -1 && cmp(r2, l1) != -1;
}
int inmid(db k1, db k2, db k3) {return sign(k1 - k3) * sign(k2 - k3) <= 0;}//k3 in [k1,k2]?1:0
struct Point {
    db x, y;
    Point operator + (const Point & a)const {return Point{a.x + x, a.y + y};}
    Point operator - (const Point & a)const {return Point{x - a.x, y - a.y};}
    Point operator * (db a) const {return Point{x * a, y * a};}
    Point operator / (db a) const {return Point{x / a, y / a};}
    bool operator < (const Point p) const {int a = cmp(x, p.x); if (a) return a == -1; return cmp(y, p.y) == -1;}
    bool operator == (const Point & a) const {return cmp(x, a.x) == 0 && cmp(y, a.y) == 0;}
    db abs() {return sqrt(x * x + y * y);}
    db dis(Point p) {return ((*this) - p).abs();}
    int getP() const {return sign(y) == 1 || (sign(y) == 0 && sign(x) == -1);}
    void input() {scanf("%lf%lf", &x, &y);}
};
db cross(Point p1, Point p2) {return p1.x * p2.y - p1.y * p2.x;}
db cross(Point p0, Point p1, Point p2) {return cross(p1 - p0, p2 - p0);}
db dot(Point p1, Point p2) {return p1.x * p2.x + p1.y * p2.y;}
int inmid(Point k1, Point k2, Point k3) {return inmid(k1.x, k2.x, k3.x) && inmid(k1.y, k2.y, k3.y);}
bool compareangle(Point p1, Point p2) {//Polar Angle Sort
    return p1.getP() < p2.getP() || (p1.getP() == p2.getP() && sign(cross(p1, p2)) > 0);
}
int clockwise(Point p1, Point p2, Point p3) { // k1 k2 k3 anticlockwise:1 clockwise:-1 others:0
    return sign(cross(p1, p2, p3));
}
struct Line {
    Point s, e;
    void input() {scanf("%lf%lf%lf%lf", &s.x, &s.y, &e.x, &e.y);}
    Point vec() {return e - s;}
    db length() {return sqrt(dot(s - e, s - e));}
    db length2() {return dot(s - e, s - e);}
};
int onS(Line l, Point p) {// On Seg?
    return inmid(l.s, l.e, p) && sign(cross(l.s - p, l.e - l.s)) == 0;
}
bool checkLL(Line l1, Line l2) {
    return cmp(cross(l1.s, l2.s, l2.e), cross(l1.e, l2.s, l2.e)) != 0;
}
bool checkLS(Line l1, Line l2) {//Intersection of Line l1 and Seg l2?
    return sign(cross(l2.s, l1.s, l1.e)) * sign(cross(l2.e, l1.s, l1.e)) <= 0;
}
int checkSS(Line l1, Line l2) {//Intersection of Two Seg?1:0
    return intersect(l1.s.x, l1.e.x, l2.s.x, l2.e.x) && intersect(l1.s.y, l1.e.y, l2.s.y, l2.e.y) && checkLS(l1, l2) && checkLS(l2, l1);
}
Point project(Line l, Point p) {
    return l.s + l.vec() * dot(p - l.s, l.vec()) / l.length2();
}
Point reflect(Line l, Point p) {//Mirror Point
    return project(l, p) * 2 - p;
}
Point getLL(Line l1, Line l2) {//Intersection Point of Line l1,l2
    db w1 = cross(l2.s, l1.s, l2.e), w2 = cross(l2.s, l2.e, l1.e); return (l1.s * w2 + l1.e * w1) / (w1 + w2);
}
db disSP(Line l, Point p) {
    Point p2 = project(l, p);
    if (inmid(l.s, l.e, p2)) return p.dis(p2); else return min(p.dis(l.s), p.dis(l.e));
}
db disSS(Line l1, Line l2) {
    if (checkSS(l1, l2)) return 0;
    return min(min(disSP(l1, l2.s), disSP(l1, l2.e)), min(disSP(l2, l1.s), disSP(l2, l1.e)));
}
db area(vector<Point> A) {//Anticlockwise
    db ans = 0;
    for (int i = 0; i < A.size(); i++) ans += cross(A[i], A[(i + 1) % A.size()]);
    return ans / 2;
}
int contain(vector<Point>A, Point p) {//2:in 1:on 0:out
    int ans = 0; A.push_back(A[0]);
    for (int i = 1; i < A.size(); i++) {
        Line l = {A[i - 1], A[i]};
        if (onS(l, p)) return 1; if (cmp(l.s.y, l.e.y) > 0) swap(l.s, l.e);
        if (cmp(l.s.y, p.y) >= 0 || cmp(l.e.y, p.y) < 0) continue;
        if (sign(cross(l.e, l.s, p)) < 0)ans ^= 1;
    }
    return ans << 1;
}
bool checkconvex(vector<Point>A) { //anticlock
    int n = A.size(); A.push_back(A[0]); A.push_back(A[1]);
    for (int i = 0; i < n; i++) if (sign(cross(A[i], A[i + 1], A[i + 2])) == -1) return 0;
    return 1;
}
vector<Point> ConvexHull(vector<Point>A, int flag = 1) { // flag=0 不严格 flag=1 严格
    int n = A.size(); vector<Point>ans(n * 2);
    sort(A.begin(), A.end()); int now = 0; if (n <= 1)return A;
    for (int i = 0; i < n; ans[now++] = A[i++])
        while (now > 1 && sign(cross(ans[now - 2], ans[now - 1], A[i])) < flag)--now;
    for (int i = n - 2, pre = now; i >= 0; ans[now++] = A[i--])
        while (now > pre && sign(cross(ans[now - 2], ans[now - 1], A[i])) < flag)--now;
    ans.resize(now - 1); return ans;
}
db convexDiameter(vector<Point>A) {
    int n = A.size(); if (n <= 1)return 0;
    int is = 0, js = 0; for (int k = 1; k <= n; ++k)is = A[k] < A[is] ? k : is, js = A[js] < A[k] ? k : js;
    int i = is, j = js; db ret = A[i].dis(A[j]); do {
        if (sign(cross(A[(i + 1) % n] - A[i], A[(j + 1) % n] - A[j])) >= 0)(++j) %= n;
        else (++i) %= n;
        ret = max(ret, A[i].dis(A[j]));
    } while (i != is || j != js);
    return ret;
}
int main() {
    int n; while (cin >> n) {
        vector<Point> A(n);
        for (int i = 0; i < A.size(); ++i)A[i].input();
        A = ConvexHull(A);
        db ans = convexDiameter(A);
        printf("%.10lf\n", ans);
    }
}