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 ll = long long; using ld = long double; using ull = unsigned long long; using uint = unsigned int;
using vi  = vector<int>; using vb  = vector<bool>; using vd  = vector<double>; using vl  = vector<ll>;
using vvi = vector<vi>;  using vvb = vector<vb>;   using vvd = vector<vd>;     using vvl = vector<vl>;

#define REP(i,n) for(ll i=0; i<(n); ++i)
#define FOR(i,b,n) for(ll i=(b); i<(n); ++i)
#define ALL(v) (v).begin(), (v).end()
#define TEN(x) ((ll)1e##x)

template<typename T> inline string join(const vector<T>& vec, string sep = " ") { stringstream ss; REP(i, vec.size()) ss << vec[i] << ( i+1 == vec.size() ? "" : sep ); return ss.str(); }


///////////////

#define EPS (1e-10)
#define NEXT(x, i) (x[(i + 1) % x.size()])

template<class T> using CR = const T &;
using P = complex<ld>;
using G = vector<P>;

int sgn(ld a, ld b = 0) {
	if (a > b + EPS) return 1;
	if (a < b - EPS) return -1;
	return 0;
}

ld dot(P a, P b) { return real(conj(a)*b); }
ld cross(P a, P b) { return imag(conj(a)*b); }

int ccw(P a, P b, P c) {
	b -= a;
	c -= a;
	if (sgn(cross(b, c))) return sgn(cross(b, c)); // clockwise(-1) or counter clockwise(1)
	if (sgn(dot(b, c)) == -1) return 2;            // c--a--b
	if (sgn(norm(b), norm(c)) == -1) return -2;    // a--b--c
	return 0;                                      // a--c--b
}

namespace std {
	bool operator < (P a, P b) {
		return sgn(real(a), real(b)) ? real(a) < real(b) : imag(a) < imag(b);
	}
}

pair<ll, ll> convex_diameter(CR<G> g) {
	ll n = g.size();

	ll is = 0, js = 0;
	FOR(i, 1, n) {
		if (imag(g[i]) > imag(g[is])) is = i;
		if (imag(g[i]) < imag(g[js])) js = i;
	}
	ld maxd = norm(g[is] - g[js]);

	ll i, maxi, j, maxj;
	i = maxi = is;
	j = maxj = js;
	do {
		if (cross(g[(i + 1) % g.size()] - g[i], g[(j + 1) % g.size()] - g[j]) >= 0) {
			j = (j + 1) % n;
		} else {
			i = (i + 1) % n;
		}
		if (norm(g[i] - g[j]) > maxd) {
			maxd = norm(g[i] - g[j]);
			maxi = i; maxj = j;
		}
	} while (i != is || j != js);
	return {maxi, maxj};
}

//////////////

int main() {
#ifdef INPUT_FROM_FILE
	ifstream cin("sample.in");
	ofstream cout("sample.out");
#endif
	cin.tie(0);
	ios_base::sync_with_stdio(false);
	cout << fixed << setprecision(30);
	ll n; cin >> n;

	G g(n);
	REP(i,n){
		ld x, y;
		cin >> x >> y;
		g[i] = { x, y };
	}
	auto p = convex_diameter(g);

	cout << abs(g[p.first] - g[p.second]) << endl;
	return 0;
}