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 C (judge verdict: Time Limit Exceeded)

/*

連立方程式「ax+by=c, dx+ey=f」の解を求めよ。

結果：Time limit exceeded

連立方程式を解くのに苦労した。
タイムリミットと言われても、どうしたものか。

*/
#include <stdio.h>
#include <math.h>

double yosute00(double n, int i);

int main()
{
	double a,b,c,d,e,f,x,y;
	int i=0,count=0;
	double sp;
	int cal,ret;
	while(1){
		ret=scanf("%lf %lf %lf %lf %lf %lf",&a,&b,&c,&d,&e,&f);
		if(ret==EOF){
			break;
		}
		while(1){
			if(a==d){
				a-=d;
				b-=e;
				c-=f;
				break;
			}
			else{
				sp=a/d;
				d*=sp;
				e*=sp;
				f*=sp;
			}
		}
		sp=c/b;
		y=yosute00(sp,4);
		a+=d;
		b+=e;
		c+=f;
		sp=(c-b*y)/a;
		x=yosute00(sp,4);
		printf("%f %f\n",x,y);
	}
	return 0;
}

double yosute00(double n, int i)
{
	double result=n;
	if(i==1){
		n+=0.5;
		n=floor(n);
	}
	else if(i==2){
		n*=10;
		n+=0.5;
		n=floor(n);
		n/=10;
	}
	else if(i==3){
		n*=100;
		n+=0.5;
		n=floor(n);
		n/=100;
	}
	else if(i==4){
		n*=1000;
		n+=0.5;
		n=floor(n);
		n/=1000;
	}
	else if(i==5){
		n*=10000;
		n+=0.5;
		n=floor(n);
		n/=10000;
	}
	return n;
}