L - Points on Cycle



x'=xcosa-ysina, y'=xsina+ycosa  (a是旋转角度)


There is a cycle with its center on the origin. 
Now give you a point on the cycle, you are to find out the other two points on it, to maximize the sum of the distance between each other 
you may assume that the radius of the cycle will not exceed 1000.


There are T test cases, in each case there are 2 decimal number representing the coordinate of the given point.


For each testcase you are supposed to output the coordinates of both of the unknow points by 3 decimal places of precision 
Alway output the lower one first(with a smaller Y-coordinate value), if they have the same Y value output the one with a smaller X. 

when output, if the absolute difference between the coordinate values X1 and X2 is smaller than 0.0005, we assume they are equal.

Sample Input

1.500 2.000
563.585 1.251

Sample Output

0.982 -2.299 -2.482 0.299
-280.709 -488.704 -282.876 487.453


#include <iostream>
#include <cmath>
#include <cstdio>
using namespace std;
class Point
    double x;
    double y;
    Point(double px, double py):x(px),y(py){}
    void Rotate(double rad);
    void Print();
void Point::Rotate(double rad)
    double tx,ty;
void Point::Print()
    printf("%.3lf %.3lf", x, y);
const double PI=3.141592654;
int main()
    int n;
    double x,y;
        Point point1(x,y);
        Point point2(x,y);
        int maxPoint=(point1.y<point2.y || point1.y==point2.y && point1.x<point2.x)?1:2;
            cout<<" ";
            cout<<" ";
    return 0;