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;

using ld = long double;
const ld eps = 1e-10, pi = acos(-1.0);

bool eq(ld a, ld b) {
	return std::abs(a - b) < eps;
}

using point = complex<ld>;

ld dot(point a, point b) {
	return real(conj(a) * b);
}

ld cross(point a, point b) {
	return imag(conj(a) * b);
}

int ccw(point a, point b, point c) {
	b -= a, c -= a;
	if (cross(b, c) > eps) return 1;	// a,b,c counter-clockwise
	if (cross(b, c) < -eps) return -1;	// a,b,c clockwise
	if (dot(b, c) < 0) return 2;		// c,a,b on a line
	if (norm(b) < norm(c)) return -2;	// a,b,c on a line
	return 0;							// a,c,b on a line
}

struct line {
	point a, b;
	line() : a(), b() {}
	line(const point& a_, const point &b_)
		: a(a_), b(b_) {}
};

bool isis_ll(line l, line m) {
	return std::abs(cross(l.b - l.a, m.b - m.a)) > eps;
}

bool isis_ls(line l, line s) {
	return (cross(l.b - l.a, s.a - l.a) * cross(l.b - l.a, s.b - l.a) < eps);
}

bool isis_lp(line l, point p) {
	return std::abs(cross(l.b - p, l.a - p)) < eps;
}

bool isis_sp(line s, point p) {
	return std::abs(s.a - p) + std::abs(s.b - p) - std::abs(s.b - s.a) < eps;
}

point proj(line l, point p) {
	ld t = dot(p - l.a, l.a - l.b) / norm(l.a - l.b);
	return l.a + t * (l.a - l.b);
}

point is_ll(line s, line t) {
	point sv = s.b - s.a, tv = t.b - t.a;
	assert(cross(sv, tv) != 0);
	return s.a + sv * cross(tv, t.a - s.a) / cross(tv, sv);
}

bool isis_ss(line s, line t) {
	if (isis_ll(s, t)) return isis_ls(s, t) && isis_ls(t, s);
	return isis_sp(s, t.a) || isis_sp(s, t.b) || isis_sp(t, s.a);
}

ld dist_lp(line l, point p) {
	return std::abs(p - proj(l, p));
}

ld dist_ll(line l, line m) {
	return isis_ll(l, m) ? 0 : dist_lp(l, m.a);
}

ld dist_ls(line l, line s) {
	return isis_ls(l, s) ? 0 : min(dist_lp(l, s.a), dist_lp(l, s.b));
}

ld dist_sp(line s, point p) {
	point r = proj(s, p);
	return isis_sp(s, r) ? std::abs(r - p) : min(std::abs(s.a - p), std::abs(s.b - p));
}

ld dist_ss(line s, line t) {
	if (isis_ss(s, t)) return 0;
	return min({ dist_sp(s, t.a), dist_sp(s, t.b), dist_sp(t, s.a), dist_sp(t, s.b) });
}

ld convex_diameter(const vector<point>& ps) {
	const int n = ps.size();
	ld d = 0;
	for (int i = 0, j = 0; i < n; ++i) {
		const point& p = ps[i];
		for (int k = (j + 1) % n; abs(ps[k] - p) > abs(ps[j] - p); k = (j + 1) % n) {
			j = k;
		}
		d = max(d, abs(ps[j] - p));
	}
	return d;
}

int main()
{
	cout << fixed << setprecision(8);
	int n;
	cin >> n;
	vector<point> ps;
	for (int i = 0; i < n; i++) {
		ld x, y;
		cin >> x >> y;
		ps.emplace_back(x, y);
	}
	cout << convex_diameter(ps) << endl;
	return 0;
}