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 <stdio.h>
#include <string.h>
#include <algorithm>
#include <iostream>
#include <math.h>
#include <assert.h>
#include <vector>
#include <queue>
#include <string>
#include <map>
#include <set>

using namespace std;
typedef long long ll;
typedef unsigned int uint;
typedef unsigned long long ull;
static const long double EPS = 1e-9;
static const long double PI = acos(-1.0);

#define REP(i, n) for (int i = 0; i < (int)(n); i++)
#define FOR(i, s, n) for (int i = (s); i < (int)(n); i++)
#define FOREQ(i, s, n) for (int i = (s); i <= (int)(n); i++)
#define FORIT(it, c) for (__typeof((c).begin())it = (c).begin(); it != (c).end(); it++)
#define MEMSET(v, h) memset((v), h, sizeof(v))

#include <complex>

typedef complex<long double> Point;
typedef vector<Point> Polygon;

static const long double INF = 1e+10;

#define CURR(P, i) (P[i])
#define NEXT(P, i) (P[(i + 1) % P.size()])

struct Line : public vector<Point> {
  Line() {;}
  Line(Point a, Point b) { push_back(a); push_back(b); }
};

namespace std {
  bool operator<(const Point &lhs, const Point &rhs) {
    return lhs.real() == rhs.real() ? lhs.imag() < rhs.imag() : lhs.real() < rhs.real();
  }
}

inline long double cross(const Point &a, const Point &b) {
  return imag(conj(a) * b);
}

inline long double dot(const Point &a, const Point &b) {
  return real(conj(a) * b);
}

inline int ccw(Point a, Point b, Point c) {
  b -= a;
  c -= a;
  if (cross(b, c) > 0) { return 1; }
  if (cross(b, c) < 0) { return -1; }
  if (dot(b, c) < 0) { return 2; }
  if (norm(b) < norm(c)) { return -2; }
  return 0;
}


Point crosspointSS(const Line &l, const Line &m) {
  long double A = cross(l[1] - l[0], m[1] - m[0]);
  long double B = cross(l[1] - l[0], l[1] - m[0]);
  if (abs(A) < EPS && abs(B) < EPS) {
    assert(false);
    return m[0];
  }
  if (abs(A) < EPS) { assert(false); }
  return m[0] + B / A * (m[1] - m[0]);
}

Polygon ConvexCut(const Polygon &P, const Line &l) {
  Polygon Q;
  for (int i = 0; i < (int)P.size(); i++) {
    Point A = CURR(P, i), B = NEXT(P, i);
    if (ccw(l[0], l[1], A) != -1) { Q.push_back(A); }
    if (ccw(l[0], l[1], A) * ccw(l[0], l[1], B) < 0) {
      Q.push_back(crosspointSS(Line(A, B), l));
    }
  }
  return Q;
}

long double Area(const Polygon &p) {
  long double ret = 0.0;
  REP(i, p.size()) {
    ret += cross(CURR(p, i), NEXT(p, i));
  }
  return ret / 2.0;
}

int n;
Polygon poly;

long double CutArea(long double theta) {
  Point vect = Point(cos(theta), sin(theta));
  Line l = Line(vect * -1000.0L, vect * 1000.0L);
  Polygon ps = ConvexCut(poly, l);
  return Area(ps);
}

int main() {
  while (scanf("%d", &n), n) {
    poly.clear();
    REP(i, n) {
      long double x, y;
      scanf("%Lf %Lf", &x, &y);
      poly.push_back(Point(x, y));
    }
    long double all = Area(poly);
    long double left = 0.0;
    long double right = PI;
    bool s = (CutArea(left) < all / 2.0);
    REP(iter, 200) {
      long double mid = (left + right) / 2.0;
      if ((CutArea(mid) < all / 2.0) == s) {
        left = mid;
      } else {
        right = mid;
      }
    }
    printf("%.16Lf %.16Lf\n", 10 * cos(left) + 1e-17, 10 * sin(left) + 1e-17);
  }
}