Problem: Sha, Nero, Eri, and Ko, who entered the University of Aizu Elementary School (Aizu University and Small), decided to participate in a programming contest called IPPC in order to play an active role as a competition programmer. However, IPPCs are required to participate in the contest as a team of three people, and it is not possible to form a team of four people. Therefore, they decided to form a team of two people and participate in a programming contest called IPOCH where two people can participate in one group.

Although the teams have been separated, their abilities are competing and they are good practice partners for each other.

One day their coach bought a whole cake to put in. They decided to eat it at Sha's house. However, when I got home and opened the box, the cake was deformed, and when viewed from above, it looked like a convex polygon instead of a circle. It seems that it was bad to have Nero carry it. There is a strawberry on top of the cake. For the time being, it was decided to cut this cake with a knife in half along a straight line through the strawberries in order to divide it into two teams. It is against their aesthetic sense to take the strawberries first before cutting the cake.

Your job is to divide a convex polygon K given on a two-dimensional plane by a straight line L passing through the origin (0, 0), and then divide the straight line L so that the areas of the two convex polygons created by the division are equal. To determine. If there are more than one, one of them will do.
<image>

Constraints

* All inputs are integers
* 3 ≤ N ≤ 20
* -100 ≤ xi ≤ 100 (1 ≤ i ≤ N)
* -100 ≤ yi ≤ 100 (1 ≤ i ≤ N)
* The number of test cases does not exceed 1,000.

Input

The input consists of multiple test cases. Each test case follows the format below.


N
x1 y1
x2 y2
...
xN yN


N is the number of vertices of the given convex polygon K. The points (xi, yi) (1 ≤ i ≤ N) are all the coordinates of the vertices that make up the convex polygon K and are different from each other. These are given in counterclockwise order. Also, a point on any side of the convex polygon K has a distance of 1 or more from the origin, and the convex polygon K always contains the origin. The end of input is indicated by one line containing one 0. Note that the three different points given by the input, including the origin, can be collinear.

Output

Output the point A (Ax, Ay) on the straight line L in the following format. However, the distance between point A and the origin must be 1 or more.


Ax Ay


When the convex polygon K is divided by the origin and a straight line passing through this point A, the difference in area between the two convex polygons must be less than 10-5. I want the coordinates of point A to be output 15 digits after the decimal point.

Example

Input

4
-100 -100
100 -100
100 100
-100 100
4
-99 -100
100 -100
100 100
-99 100
4
-100 -99
100 -99
100 100
-100 100
14
-99 -70
-92 -79
10 -98
37 -100
62 -95
77 -69
88 -47
92 -10
96 28
100 91
42 92
-62 92
-88 91
-98 64
0


Output

100.000000000000000 0.000000000000000
100.000000000000000 0.000000000000000
0.000000000000000 100.000000000000000
-96.983291994836122 66.745111613942484

Solution (in CPP):

Solution:
#include <algorithm>
#include <bitset>
#include <cassert>
#include <cctype>
#include <climits>
#include <cmath>
#include <complex>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <ctime>
#include <deque>
#include <functional>
#include <iostream>
#include <iterator>
#include <list>
#include <map>
#include <numeric>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <utility>
#include <vector>
using namespace std;

#define dump(a) (cerr << (#a) << " = " << (a) << endl)
#define FOR(it,c) for(__typeof((c).begin())it=(c).begin(); it!=(c).end();++it)
#define RFOR(it,c) for(__typeof((c).rbegin())it=(c).rbegin(); it!=(c).rend();++it)
#define exist(c, v) (find((c).begin(), (c).end(), (v)) != (c).end())

template<class T> inline void chmax(T& a, const T& b) { if(b > a) a = b; }
template<class T> inline void chmin(T& a, const T& b) { if(b < a) a = b; }

template<typename T, typename U> ostream& operator<<(ostream& os, const pair<T, U>& p) {
	os << '(' << p.first << ", " << p.second << ')';
	return os;
}

template<typename T> ostream& operator<<(ostream& os, const vector<T>& v) {
	copy(v.begin(), v.end(), ostream_iterator<T>(os, " "));
	return os;
}

const double EPS = 1e-9;

struct point {
	double x, y;
	point():x(0), y(0) {}
	point(double x, double y):x(x), y(y) {}
	point(const point& p):x(p.x), y(p.y) {}

	point operator+ (const point& p) const {
		return point(x + p.x, y + p.y);
	}

	point operator- (const point& p) const {
		return point(x - p.x, y - p.y);
	}

	point operator* (const double s) const {
		return point(x * s, y * s);
	}

	point operator* (const point& p) const {
		return point(x * p.x - y * p.y, x * p.y + y * p.x);
	}

	point operator/ (const double s) const {
		return point(x / s, y / s);
	}

	bool operator< (const point& p) const {
		return x + EPS < p.x || abs(x - p.x) < EPS && y + EPS < p.y;
	}

	bool operator== (const point& p) const {
		return abs(x - p.x) < EPS && abs(y - p.y) < EPS;
	}
};

inline point rotate90(const point& p) {
	return point(-p.y, p.x);
}

inline point rotate(const point& p, const double theta) {
	const double s = sin(theta), c = cos(theta);
	return point(c * p.x - s * p.y, s * p.x + c * p.y);
}

inline double angle(const point& p) {
	return atan2(p.y, p.x);
}

inline double abs(const point& p) {
	return sqrt(p.x * p.x + p.y * p.y);
}

inline double norm(const point& p) {
	return p.x * p.x + p.y * p.y;
}

inline double dot(const point& l, const point& r) {
	return l.x * r.x + l.y * r.y;
}

inline double cross(const point& l, const point& r) {
	return l.x * r.y - l.y * r.x;
}

struct line {
	point a, b;
	line(point a, point b):a(a), b(b){}
};

struct segment {
	point a, b;
	segment(point a, point b):a(a), b(b){}
};

struct circle {
	point c;
	double r;
	circle(point c, double r):c(c), r(r){}
};

typedef vector<point> polygon;

inline int ccw(const point& a, point b, point c) {
	b = b - a;
	c = c - a;
	const double tmp = cross(b, c);
	if(tmp > EPS) return 1; // ccw
	if(tmp < -EPS) return -1; // cw
	if(dot(b, c) < 0) return 2; // c, a, b 順に一直線上
	if(norm(b) < norm(c)) return -2; // a, b, c 順に一直線上
	return 0; //a, c, b 順で一直線上
}

inline point projection(const line& l, const point& p) {
	const point dif = l.b - l.a;
	const double tmp = dot(p - l.a, dif) / norm(dif);
	return l.a + dif * tmp;
}

inline bool intersect(const line& l, const line& m) {
	return abs(cross(l.b - l.a, m.b - m.a)) > EPS || abs(cross(l.b - l.a, m.b - l.a)) < EPS;
}

inline bool intersect(const line& l, const segment& s) {
	return cross(l.b - l.a, s.a - l.a) * cross(l.b - l.a, s.b - l.a) < EPS;
}

inline bool intersect(const line& l, const point& p) {
	return abs(ccw(l.a, l.b, p)) != -1;
}

inline bool intersect(const segment& s, const segment& t) {
	return ccw(s.a, s.b, t.a) * ccw(s.a, s.b, t.b) <= 0 &&
		ccw(t.a, t.b, s.a) * ccw(t.a, t.b, s.b) <= 0;
}

inline bool intersect(const segment& s, const point& p) {
	return ccw(s.a, s.b, p) == 0;
}

inline bool intersect(const circle& c, const point& p) {
	return abs(c.c - p) <= c.r + EPS;
}

inline bool intersect(const circle& c, const circle& d) {
	return abs(c.c - d.c) <= c.r + d.r && abs(c.c - d.c) >= abs(c.r - d.r);
}

inline double dist(const line& l, const point& p) {
	return abs(cross((l.b - l.a), (p - l.a))) / abs(l.b - l.a);
}

inline double dist(const line& l, const line& m) {
	return intersect(l, m) ? 0 : dist(l, m.a);
}

inline double dist(const line& l, const segment& s) {
	return intersect(l, s) ? 0 : min(dist(l, s.a), dist(l, s.b));
}

inline double dist(const segment& s, const point& p) {
	if(dot((s.b - s.a), (p - s.a)) < 0) return abs(p - s.a);
	if(dot((s.a - s.b), (p - s.b)) < 0) return abs(p - s.b);
	return dist(line(s.a, s.b), p);
}

inline double dist(const segment& s, const segment& t) {
	if(intersect(s, t))
		return 0;
	return min(min(dist(s, t.a), dist(s, t.b)), min(dist(t, s.a), dist(t, s.b)));
}

inline double dist(const point& a, const point& b) {
	return abs(a - b);
}

inline bool intersect(const circle& c, const segment& s) {
	return dist(s, c.c) < c.r + EPS;
}

inline point crosspoint(const line& l, const line& m) {
	const double tmp = cross(l.b - l.a, m.b - m.a);
	if(abs(tmp) < EPS) // 平行
		return l.a;
	return m.a + (m.b - m.a) * cross(l.b - l.a, l.a - m.a) * (1.0 / tmp);
}

inline point crosspoint(const segment& s, const segment& t) {
	if(!intersect(s, t)) // 交点を持たない
		return s.a; // 用改善

	const double tmp = cross(s.b - s.a, t.b - t.a);
	if(abs(tmp) < EPS) { // 一直線上
		if(intersect(s, t.a)) return t.a;
		if(intersect(s, t.b)) return t.b;
		if(intersect(t, s.a)) return s.a;
		return s.b;
	}

	return t.a + (t.b - t.a) * cross(s.b - s.a, s.b - t.a) * (1.0 / tmp);
}

//先に，intersectを用いて交差判定を行うこと
inline vector<point> crosspoint(const circle &c, const circle& d) {
	vector<point> res;
	if(abs(c.c - d.c) < EPS) // 中心の座標が同じ
		return res;

	const double tmp = abs(c.c - d.c);
	const double rc = (tmp * tmp + c.r * c.r - d.r * d.r) / (tmp + tmp);
	const double rs = sqrt(c.r * c.r - rc * rc);
	const point diff = (d.c - c.c) / tmp;
	res.push_back(point(c.c + diff * point(rc, rs)));
	res.push_back(point(c.c + diff * point(rc, -rs)));
	return res;
}

inline vector<point> crosspoint(const circle& c, const line& l) {
	vector<point> res;
	const point h = projection(l, c.c);
	const double d = abs(h - c.c);
	if(d > c.r - EPS) {
		res.push_back(h);
	}
	else if(d <= c.r - EPS) {
		point v = l.b - l.a;
		v = v * sqrt(c.r * c.r - d * d) / abs(v);
		res.push_back(h + v);
		res.push_back(h - v);
	}

	return res;
}

inline vector<point> crosspoint(const circle& c, const segment& s) {
	vector<point> res;
	vector<point> tmp = crosspoint(c, line(s.a, s.b));
	for(int i = 0; i < tmp.size(); ++i)
		if(intersect(s, tmp[i]))
			res.push_back(tmp[i]);

	return res;
}

inline double areaTriangle(point a, point b, const point& c) {
	a = a - c;
	b = b - c;
	return fabs(a.x * b.y - b.x * a.y) / 2.0;
}

inline double area(const polygon& p) {
	const int num = p.size();
	if(num < 3)
		return 0;

	if(num == 3)
		return areaTriangle(p[0], p[1], p[2]);

	double res = 0.0;
	for(int i = 0; i < num; ++i)
		res += cross(p[i], p[(i + 1) % num]);

	return abs(res) * 0.5;
}

// L.aからL.bの方向を見た場合に，点aが左側に来る．
inline line bisector(const point& a, const point& b) {
	return line(point((a.x - a.y + b.x + b.y) / 2.0, (a.y + a.x + b.y - b.x) / 2.0),
				point((a.x + a.y + b.x - b.y) / 2.0, (a.y - a.x + b.y + b.x) / 2.0));
}

// L.aからL.bを向いた時の左側を残して切断する．
inline polygon convex_cut(const polygon& p, const line& l) {
	const int num = p.size();
	polygon res;
	for(int i = 0; i < num; ++i) {
		const int next = (i + 1) % num;
		const int tmp = ccw(l.a, l.b, p[i]);
		if(tmp != -1)
			res.push_back(p[i]);

		if(tmp * ccw(l.a, l.b, p[next]) == -1)
			res.push_back(crosspoint(l, line(p[i], p[next])));
	}

	return res;
}

vector<point> tangent(const circle& c, const point& p) {
	const double x = norm(p - c.c);
	double d = x - c.r * c.r;
	if(d < -EPS) return vector<point>();
	d = max(d, 0.0);
	const point q1 = (p - c.c) * (c.r * c.r / x);
	const point q2 = rotate90((p - c.c) * (-c.r * sqrt(d) / x));
	vector<point> res;
	res.push_back(c.c + q1 - q2);
	res.push_back(c.c + q1 + q2);
	return res;
}

vector<line> tangent(const circle &a, const circle& b) {
	vector<line> res;
	if(abs(a.r - b.r) < EPS) {
		point dir = b.c - a.c;
		dir = rotate90(dir * (a.r / abs(dir)));
		res.push_back(line(a.c + dir, b.c + dir));
		res.push_back(line(a.c - dir, b.c - dir));
	}
	else {
		point p = a.c * (-b.r) + b.c * a.r;
		p = p * (1.0 / (a.r - b.r));
		vector<point> ps = tangent(a, p), qs = tangent(b, p);
		for(int i = 0; i < min<int>(ps.size(), qs.size()); ++i)
			res.push_back(line(ps[i], qs[i]));
	}

	point p = a.c * b.r + b.c * a.r;
	p = p * (1.0 / (a.r + b.r));
	vector<point> ps = tangent(a, p), qs = tangent(b, p);
	for(int i = 0; i < min<int>(ps.size(), qs.size()); ++i)
		res.push_back(line(ps[i], qs[i]));

	return res;
}

// 三角形と点の包含関係，境界は含まない．
// 含める場合は == -1を <= 0とする．
bool contain_triangle(const polygon& triangle, const point& p) {
	const int d1 = ccw(p, triangle[0], triangle[1]), d2 = ccw(p, triangle[1], triangle[2]), d3 = ccw(p, triangle[2], triangle[0]);
	return !(d1 * d2 == -1 || d2 * d3 == -1 || d3 * d1 == -1);
}

// 多角形と点の包含関係．境界を含む．
bool contain(const polygon& P, const point& p) {
	const int n = P.size();
	bool res = false;
	for(int i = 0; i < n; ++i) {
		point v1 = P[i] - p, v2 = P[(i + 1) % n] - p;
		if(v1.y > v2.y)
			swap(v1, v2);

		if(v1.y < EPS && EPS < v2.y && cross(v1, v2) > EPS)
			res = !res;

		if(abs(cross(v1, v2)) < EPS && dot(v1, v2) < EPS)
			return true; // 境界を含まないなら，falseにする．
	}

	return res;
}


int main() {
	const point o(0, 0);
	for(int n; scanf("%d", &n), n;) {
		polygon cake;
		cake.reserve(n);
		for(int i = 0; i < n; ++i) {
			int x, y;
			scanf("%d %d", &x, &y);
			cake.push_back(point(x, y));
		}

		const double s = area(cake);

		double low = 0.0, high = M_PI;
		if(area(convex_cut(cake, line(o, point(1.0, 0.0)))) > s / 2.0) {
			low = -M_PI;
			high = 0;
		}

		for(int i = 0; i < 100; ++i) {
			const double mid = (low + high) / 2.0;
			if(area(convex_cut(cake, line(o, point(cos(mid), sin(mid))))) < s / 2.0)
				low = mid;
			else
				high = mid;
		}

		const double ans = (low + high) / 2.0;
		printf("%.15lf %.15lf\n", 100 * cos(ans), 100 * sin(ans));
	}

	return EXIT_SUCCESS;
}