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.
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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 <iostream>
#include <algorithm>
#include <vector>
#include <cmath>
using namespace std;
#define DEBUG(x) cerr << #x << " = " << x << endl
typedef double D;
const D EPS = 1e-8;
const D PI = M_PI;
struct P {
  D x, y;
  P() { }
  P(D x_, D y_) : x(x_), y(y_) { }
};
struct L {
  P a, b;
  L() { }
  L(P a_, P b_) : a(a_), b(b_) { }
};
struct C {
  P p; D r;
  C() { }
  C(P p_, D r_) : p(p_), r(r_) { }
};
P operator +(P a, P b) { return P(a.x + b.x, a.y + b.y); }
P operator -(P a, P b) { return P(a.x - b.x, a.y - b.y); }
P operator *(P a, D b) { return P(a.x * b, a.y * b); }
P operator /(P a, D b) { return P(a.x / b, a.y / b); }
P vec(P from, P to) { return to - from; }
D inp(P a, P b) { return a.x*b.x + a.y*b.y; }
D outp(P a, P b) { return a.x*b.y - a.y*b.x; }
D norm(P p) { return inp(p, p); }
D abs(P p) { return sqrt(norm(p)); }
int sig(D a, D b) {
  if(a < b - EPS) return -1;
  if(a > b + EPS) return +1;
  return 0;
}
// !!! とりあえずここまで写してください !!!
D arg(P p) { return atan2(p.y, p.x); }
P rot90(P p) { return P(-p.y, p.x); }
P rot(P p, D radian) {
  P q;
  q.x = cos(radian)*p.x - sin(radian)*p.y;
  q.y = sin(radian)*p.x + cos(radian)*p.y;
  return q;
}
int ccw(P a, P b, P c) { // 重なっている点があるとうまく動かないと思われる
  b = vec(a, b); c = vec(a, c);

  // a - b - c が折れ曲がるとき
  if(sig(outp(b, c), 0.0) > 0) return +1; // 反時計回り
  if(sig(outp(b, c), 0.0) < 0) return -1; // 時計回り

  // a - b - c が直線上に並ぶとき
  if(sig(inp(b, c), 0.0) < 0) return +2; // c - a - b
  if(norm(b) < norm(c))       return -2; // a - b - c
  return 0;                              // a - c - b
}
P projection(L l, P p) { // 直線lに対する点pの写像
  P a = vec(l.a, l.b);
  P b = vec(l.a, p);
  D t = inp(a, b) / norm(a);
  return l.a + a * t;
}
P reflection(L l, P p) { // 直線lに対する点pの反射
  return p + vec(p, projection(l, p)) * 2;
}

// 線分と点
bool iSP(L s, P p) {
  return ccw(s.a, s.b, p) == 0;
}
D dSP(L s, P p) {
  P r = projection(s, p);
  if(iSP(s, r)) return abs(p - r); // 写像がs上にある
  return min(abs(p - s.a), abs(p - s.b)); // 写像がs上にない
}

// 直線と点
bool iLP(L l, P p) {
  return abs(ccw(l.a, l.b, p)) != 1;
}
D dLP(L l, P p) {
  return abs(p - projection(l, p));
}

// 線分と線分
bool iSS(L s, L t) { // 端を含む
  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;
}
D dSS(L s, L t) {
  if(iSS(s, t)) return 0;
  return min(min(dSP(s, t.a), dSP(s, t.b)),
      min(dSP(t, s.a), dSP(t, s.b)));
}
// P cSS(L s, L t) はiSS(s,t)を確認してからcLLを使う

// 直線と直線
bool iLL(L l, L m) {
  return sig(outp(vec(l.a, l.b), vec(m.a, m.b)), 0.0) != 0 || // 平行でない
    sig(outp(vec(l.a, l.b), vec(l.a, m.a)), 0.0) == 0;   // 同じ直線
}
D dLL(L l, L m) {
  return iLL(l, m) ? 0.0 : dLP(l, m.a);
}
P cLL(L l, L m) {
  D d = outp(vec(m.a, m.b), vec(l.a, l.b));
  return l.a + vec(l.a, l.b) * outp(vec(m.a, m.b), vec(l.a, m.b)) / d;
}

// 直線と線分
bool iLS(L l, L s) {
  return sig(outp(vec(l.a, l.b), vec(l.a, s.a)), 0.0) *
    sig(outp(vec(l.a, l.b), vec(l.a, s.b)), 0.0) <= 0;
}
D dLS(L l, L s) {
  if(iSS(l, s)) return 0.0;
  return min(dLP(l, s.a), dLP(l, s.b));
}
// P cLS(L s, L t) はiSS(s,t)を確認してからcLLを使う

// 円と直線
int iCL(C c, L l) {
  D d = dLP(l, c.p);
  int s = sig(d, c.r);
  if(s < 0) return 2;
  if(s == 0) return 1;
  return 0;
}
vector<P> cCL(C c, L l) {
  P a = projection(l, c.p);
  D s = abs(c.p - a);
  D t = sqrt(c.r*c.r-s*s);
  P v = vec(l.a,l.b)/abs(l.a-l.b);
  vector<P> res;
  res.push_back(a+v*t);
  res.push_back(a-v*t);
  return res;
}

// 円と点
vector<P> tCP(C c, P p) {
  vector<P> res;
  D d = abs(c.p - p);
  if(sig(d,c.r) < 0) return res;
  D rc = c.r*c.r/d;
  D rs = sqrt(max(0.0, c.r*c.r - rc*rc));
  P v = (p - c.p) / abs(p - c.p);
  res.emplace_back(c.p + v*rc + rot90(v)*rs);
  res.emplace_back(c.p + v*rc - rot90(v)*rs);
  return res;
}

// 円と線分
int iCS(C c, L s) {
  if(sig(abs(c.p - s.a), c.r) <= 0 && sig(abs(c.p - s.b), c.r) <= 0)
    return -2; // 線分の両端が円の内側にある
  if(sig(abs(c.p - s.a), c.r) <= 0 || sig(abs(c.p - s.b), c.r) <= 0)
    return -1; // 線分の端の一方が円の内側、他方が円の外側にある
  if(sig(dLP(s, c.p), c.r) < 0) // dSPじゃなくてdLPで十分だと思う
    return +2; // 線分と円周が2点で交わる
  if(sig(dLP(s, c.p), c.r) == 0)
    return +1; // 線分と円周が1点で交わる
  return 0;
}
// vector<P> cCS(C c, L s) はiCSの状態によって定義が難しいが、
// p <- cCL(c,s)からiSP(s,p)==trueのものだけを抜き出せば良い気がする

// 円と円
// int iCC(C a, C b) はaがbの内側にある場合などもあるので定義が難しい
// 2点で接していることを確認すること
vector<P> cCC(C a, C b) {
  D d = abs(b.p - a.p);
  D x = (d*d + a.r*a.r - b.r*b.r) / (2*d);
  D y = sqrt(a.r*a.r - x*x);
  P v = (b.p - a.p) / d;
  vector<P> res;
  res.emplace_back(a.p + v*x + rot90(v)*y);
  res.emplace_back(a.p + v*x - rot90(v)*y);
  return res;
}
// tCCinに同じ円を食わせたり、
// tCCoutにどちらかが他方に囲まれている円を食わせたりすると破綻することは分かっているが、
// そのあたりが厳密にverifyできていない (AOJ 2201でACすることは確認した)
vector<L> tCCout(C a, C b) {
  vector<L> res;
  if(sig(abs(a.r - b.r), abs(a.p - b.p)) >= 0) return res;
  if(sig(a.r, b.r) == 0) {
    P v = (b.p - a.p) / abs(b.p - a.p);
    v = rot90(v);
    res.emplace_back(a.p + v*a.r, b.p + v*b.r);
    res.emplace_back(a.p - v*a.r, b.p - v*b.r);
    return res;
  }
  P p = (a.p*b.r - b.p*a.r) / (b.r - a.r);
  vector<P> at = tCP(a, p);
  vector<P> bt = tCP(b, p);
  for(int i = 0; i < (int)min(at.size(), bt.size()); ++i) {
    res.emplace_back(at[i], bt[i]);
  }
  return res;
}
vector<L> tCCin(C a, C b) {
  vector<L> res;
  if(sig(abs(a.r + b.r), abs(a.p - b.p)) >= 0) return res;
  P p = (a.p*b.r + b.p*a.r) / (a.r + b.r);
  vector<P> at = tCP(a, p);
  vector<P> bt = tCP(b, p);
  for(int i = 0; i < (int)min(at.size(), bt.size()); ++i) {
    res.emplace_back(at[i], bt[i]);
  }
  return res;
}
vector<L> tCC(C a, C b) {
  vector<L> res;
  for(L l : tCCout(a, b)) res.push_back(l);
  for(L l : tCCin(a, b)) res.push_back(l);
  return res;
}

// 多角形
typedef vector<P> G;
// 半時計回りを仮定している (時計回りならabsを取る)
D area(G g) {
  D res = 0.0;
  for(int i = 0; i < (int)g.size(); i++) {
    res += outp(g[i], g[(i+1)%g.size()]);
  }
  return res / 2.0;
}
// ON = 0, IN = 1, OUT = -1
int containsGP(G g, P p) {
  int side = -1;
  for(int i = 0; i < (int)g.size(); i++) {
    if(ccw(g[i], g[(i+1)%g.size()], p) == 0) return 0;
    P a = vec(p, g[i]);
    P b = vec(p, g[(i+1)%g.size()]);
    if(a.y > b.y) swap(a, b);
    if(sig(a.y, 0.0) <= 0 && sig(b.y, 0.0) > 0 && sig(outp(a, b), 0.0) > 0) side *= -1;
  }
  return side;
}
bool operator <(P a, P b) {
  if(sig(a.x, b.x) != 0) return a.x < b.x;
  return a.y < b.y;
}
// 凸包を構成する点を得る。半時計回り
// 辺上の点も含めるときは ccw(..) == -1 とすること
G convex_hull(vector<P> ps) {
  int N = ps.size();
  int k = 0; // 凸包を構成する点の数
  sort(ps.begin(), ps.end());
  G res(N * 2);
  for(int i = 0; i < N; i++) {
    // 時計回りの折れ曲がりがあったら、折れ点を削除
    while(k >= 2 && ccw(res[k - 2], res[k - 1], ps[i]) <= 0) k--;
    res[k++] = ps[i];
  }
  int t = k + 1;
  // 右端は取らない (重複するから)
  // 左端は取る (重複するけど、あとから取り除いたほうが楽)
  for(int i = N - 2; i >= 0; i--) {
    while(k >= t && ccw(res[k - 2], res[k - 1], ps[i]) <= 0) k--;
    res[k++] = ps[i];
  }
  res.resize(k - 1);
  return res;
}
// l.a -> l.b の厳密に右側の領域を切り落とす
G convex_cut(G g, L l) {
  G res;
  for(int i = 0; i < (int)g.size(); ++i) {
    P a = g[i];
    P b = g[(i+1)%g.size()];
    if(ccw(l.a, l.b, a) != -1) res.push_back(a);
    // 端の点を含まないiLS
    if(ccw(l.a, l.b, a)*ccw(l.a, l.b, b) < 0) {
      res.push_back(cLL(L(a, b), l));
    }
  }
  return res;
}
// aとbの垂直二等分線をaが左側に来るように計算する
L bisector(P a, P b) {
  P p = (a + b) / 2;
  return L(p, p + rot90(vec(a,b)));
}
// 外枠をgとして点集合vのk番目の点のボロノイ領域を返す
G voronoi_cell(G g, vector<P> v, int k) {
  for(int i = 0; i < (int)v.size(); ++i) {
    if(i == k) continue;
    g = convex_cut(g, bisector(v[k], v[i]));
  }
  return g;
}
int main() {
  while(true) {
    int N; cin >> N;
    if(N == 0) break;
    G g;
    for(int i = 0; i < N; ++i) {
      double x, y; cin >> x >> y;
      g.emplace_back(x,y);
    }
    double up = area(convex_cut(g, L(P(0.0,0.0), P(1.0,0.0))));
    double down = area(convex_cut(g, L(P(0.0,0.0), P(-1.0,0.0))));
    cout.setf(ios::fixed);
    cout.precision(15);
    double lb, ub;
    if(sig(up,down) == 0) {
      cout << 1.0 << ' ' << 0.0 << endl;
      continue;
    }

    if(sig(up,down) > 0) {
      lb = PI, ub = 2*PI;
    }
    else {
      ub = PI, lb = 0.0;
    }

    double all = area(g);

    for(int i = 0; i < 100; i++) {
      double mid = (ub + lb) / 2;
      // DEBUG(lb); DEBUG(ub);
      L l(P(0.0, 0.0), P(100*cos(mid),100*sin(mid)));
      double a = area(convex_cut(g, l));
      // DEBUG(abs(a-all));
      if(a > all / 2) ub = mid;
      else lb = mid;
    }

    cout << 100*cos(lb) << ' ' << 100*sin(lb) << endl;
  }
}