Problem:
Simultaneous Equation

Write a program which solve a simultaneous equation:

ax + by = c

dx + ey = f

The program should print x and y for given a, b, c, d, e and f (-1,000 ≤ a, b, c, d, e, f ≤ 1,000). You can suppose that given equation has a unique solution.

Input

The input consists of several data sets, 1 line for each data set. In a data set, there will be a, b, c, d, e, f separated by a single space. The input terminates with EOF.

Output

For each data set, print x and y separated by a single space. Print the solution to three places of decimals. Round off the solution to three decimal places.

Sample Input 1

1 2 3 4 5 6
2 -1 -2 -1 -1 -5

Output for the Sample Input 1

-1.000 2.000
1.000 4.000

Sample Input 2

2 -1 -3 1 -1 -3
2 -1 -3 -9 9 27

Output for the Sample Input 2

0.000 3.000
0.000 3.000

Solution in JAVA (judge verdict: Wrong Answer)

import java.util.Scanner;

public class Main
{
	public static void main(String[] args)
	{
		Scanner scan=new Scanner(System.in);
		int[] c=new int[6];
		double x,y;
		while(scan.hasNext()){
			for(int i=0;i<c.length;i++){
				c[i]=scan.nextInt();
			}
			if(c[0]==0){
				y=c[2]/(double)c[1];
				x=(c[5]-c[4]*y)/(double)c[3];
			}else if(c[1]==0){
				x=c[2]/(double)c[0];
				y=(c[5]-c[3]*x)/(double)c[4];
			}else if(c[3]==0){
				y=c[5]/(double)c[4];
				x=(c[2]-c[1]*y)/(double)c[0];
			}else if(c[4]==0){
				x=c[5]/(double)c[3];
				y=(c[2]-c[0]*x)/(double)c[1];
			}else{
				x=(c[4]*c[2]-c[1]*c[5])/(double)(c[4]*c[0]-c[1]*c[3]);
				y=(c[2]*c[3]-c[0]*c[5])/(double)(c[1]*c[3]-c[0]*c[4]);
			}
			System.out.printf("%.3f %.3f",x,y);
		}
		scan.close();
	}
}