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<string>
#include<algorithm>
#include<vector>
#include<iomanip>
#include<math.h>
#include<complex>
#include<queue>
#include<deque>
#include<map>
#include<set>
#include<bitset>
using namespace std;
#define REP(i,m,n) for(int i=(int)m ; i < (int) n ; i++ )
#define rep(i,n) REP(i,0,n)
typedef long long ll;
typedef pair<int,int> pint;
const int inf=1e9+7;
const ll longinf=1LL<<60 ;
const ll mod=1e9+7 ;
int dx[4]={1,0,-1,0} , dy[4]={0,1,0,-1} ;

const double eps=1e-6;
double add(double a,double b){
    if(abs(a+b)<eps)return 0;
    return a+b;
}

typedef pair<double,double> P;
P operator+(const P& x,const P& y){
    return {add(x.first,y.first),add(x.second,y.second)};
}

P operator-(const P& x,const P& y){
    return {add(x.first,-y.first),add(x.second,-y.second)};

}

P operator*(const P& x,const double& y){
    return {x.first*y,x.second*y};
}
double dot(P x,P y){
    return add(x.first*y.first,x.second*y.second);


}

double det(P x,P y){
    return add(x.first*y.second,-x.second*y.first);
}

bool onseg(P p1,P p2,P q){
    return det(p1-q,p2-q)==0 &&dot(p1-q,p2-q)<=0; 
}

P intersection(P p1,P p2,P q1,P q2){
    return p1+(p2-p1)*(det(q2-q1,q1-p1)/det(q2-q1,p2-p1));
}

bool cross(P p1,P p2,P q1,P q2){
    if(det(p1-p2,q1-q2)==0)return onseg(p1,p2,q1)||onseg(p1,p2,q2)||onseg(q1,q2,p1)||onseg(q1,q2,p2);
    P ret=intersection(p1,p2,q1,q2);
    return onseg(p1,p2,ret)&&onseg(q1,q2,ret);
}

double dist(P p1,P p2,P q){
    if(dot(q-p1,p2-p1)<=0)return sqrt(dot(q-p1,q-p1));
    if(dot(q-p2,p1-p2)<=0)return sqrt(dot(q-p2,q-p2));
    P d=p1-p2;
    P ret=q+P{d.second,-d.first};
    P h=intersection(p1, p2, q, ret);
    return sqrt(dot(q-h,q-h));
}
double d(P p,P q){
    return sqrt(dot(p-q,p-q));
}

double area(P p1,P p2,P p3){
    P q1=p2-p1,q2=p3-p1;
    return abs(det(q1,q2)/2);
}
double Parea(vector<P> vp){
    int m=vp.size();
    if(m<2)return 0;
    double ret=0;
    rep(i,m-2){
        ret+=area(vp[0],vp[i+1],vp[i+2]);
    }
    return ret;
}

int n;
P a[30];
double calc(double arg){
    P b={cos(arg),sin(arg)};
    b=b*300;
    while(onseg(b,b*(-1),a[0])){
        P c[n];
        rep(i,n)c[i]=a[(i+1)%n];
        rep(i,n)a[i]=c[i];
    }
    bool flag=(det(a[0],b)>0);
    vector<P> v[2];
    v[flag].push_back(a[0]);
    rep(i,n-1){
        if(cross(a[i],a[i+1],b,b*(-1))&&!onseg(b,b*(-1),a[i])){
            flag=1-flag;
            v[0].push_back(intersection(a[i], a[i+1], b, b*(-1)));
            v[1].push_back(intersection(a[i], a[i+1], b, b*(-1)));
        }
        v[flag].push_back(a[i+1]);
    }
    if(cross(a[0],a[n-1],b,b*(-1))&&!onseg(b,b*(-1),a[n-1])){
            v[0].push_back(intersection(a[0], a[n-1], b, b*(-1)));
            v[1].push_back(intersection(a[0], a[n-1], b, b*(-1)));
    }
    return Parea(v[0])-Parea(v[1]);
}

const double pi=acos(-1);
int main(){
    cout<<fixed<<setprecision(16);
    while(cin>>n,n!=0){
        rep(i,n)cin>>a[i].first>>a[i].second;
        double left=0,right=pi,mid;
        while(right-left>1e-10){
            double ll=calc(left);
            double rr=calc(right);
            mid=(left+right)/2;
            double mm=calc(mid);
            if(abs(mm)<1e-7)break;
            if(ll*mm>0)left=mid;
            else right=mid;
        }
        cout<<10*cos(mid)<<" "<<10*sin(mid)<<endl;
    }
    return 0;
}