Problem: Mr. KM, the mayor of KM city, decided to build a new elementary school. The site for the school has an awkward polygonal shape, which caused several problems. The most serious problem was that there was not enough space for a short distance racetrack. Your task is to help Mr. KM to calculate the maximum possible length for the racetrack that can be built in the site. The track can be considered as a straight line segment whose width can be ignored. The boundary of the site has a simple polygonal shape without self-intersection, and the track can touch the boundary. Note that the boundary might not be convex.



Input

The input consists of multiple test cases, followed by a line containing "0". Each test case has the following format. The first line contains an integer N (3 \leq N \leq 100). Each of the following N lines contains two integers x_i and y_i (-1,000 \leq x_i, y_i \leq 1,000), which describe the coordinates of a vertex of the polygonal border of the site, in counterclockwise order.

Output

For each test case, print its case number and the maximum possible length of the track in a line. The answer should be given as a floating point number with an absolute error of at most 10^{-6}.

Example

Input

4
0 0
10 0
10 10
0 10
3
0 0
1 0
0 1
0


Output

Case 1: 14.142135624
Case 2: 1.41421356

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 abs(a - b) < eps;
}

using Point = complex<ld>;
class Line {
public:
	Point a, b;
};
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
}

bool isis_ll(Line l, Line m) {
	return 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 abs(cross(l.b - p, l.a - p)) < eps;
}

bool isis_sp(Line s, Point p) {
	return abs(s.a - p) + abs(s.b - p) - 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 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) ? abs(r - p) : min(abs(s.a - p), 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) });
}

bool is_in_polygon(const vector<Point>& poly, const Point& p) {
	int n = poly.size();
	ld sum = 0;
	for (int i = 0; i < n; ++i) {
		Point p1 = poly[i], p2 = poly[(i + 1) % n];
		if (isis_sp((Line) { p1, p2 }, p)) return true;
		sum += arg((p2 - p) / (p1 - p));
	}
	return abs(sum) > pi;
}

ld calc(const Line& l, const vector<Point>& ps) {
	int n = ps.size();
	Point v = l.b - l.a;
	assert(abs(v) > eps);
	v /= abs(v);
	vector<ld> pos = { 0, abs(l.b - l.a) };
	for (int i = 0; i < n; i++) {
		Line s = { ps[i], ps[(i + 1) % n] };
		if (abs(cross(l.b - l.a, s.b - s.a)) > eps && isis_ss(l, s)) {
			auto pp = is_ll(l, s);
			ld x = dot(pp - l.a, v);
			if (0 < x && x < pos[1]) pos.push_back(x);
		}
	}
	sort(pos.begin(), pos.end());
	ld res = 0;
	for (int i = 1; i < (int)pos.size(); i++) {
		if (!is_in_polygon(ps, l.a + v * ((pos[i] + pos[i - 1]) / 2.0))) break;
		res = pos[i];
	}
	return res;
}

int main()
{
	ios::sync_with_stdio(false), cin.tie(0);
	cout << fixed << setprecision(10);
	int n, tcase = 0;
	while (cin >> n, n) {
		vector<Point> ps;
		for (int i = 0; i < n; i++) {
			ld x, y;
			cin >> x >> y;
			ps.emplace_back(x, y);
		}
		ld res = 0;
		for (int i = 0; i < n; i++) {
			for (int j = i + 1; j < n; j++) {
				Line l = { ps[i], ps[j] };
				if (calc(l, ps) < abs(l.a - l.b) - eps) continue;
				Point v = ps[j] - ps[i];
				assert(abs(v) > eps);
				v /= abs(v);
				Point p1 = ps[j] + v * calc((Line) { ps[j], ps[j] + v * (ld)1e5 }, ps);
				Point p2 = ps[i] - v * calc((Line) { ps[i], ps[i] - v * (ld)1e5 }, ps);
				res = max(res, abs(p1 - p2));
			}
		}
		cout << "Case " << ++tcase << ": " << res << endl;
	}
	return 0;
}