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)

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


int main(void){
  
    //int n, i;
    int mat[2][2], c, f;
    double mat2[2][2];
    double deltamat;
    
    while(scanf("%d %d %d %d %d %d", &mat[0][0], &mat[0][1], &c, &mat[1][0], &mat[1][1], &f)){
        deltamat = (double)mat[0][0] * mat[1][1] - (double)mat[0][1] * mat[1][0];
        //printf("%f\n", deltamat);
        mat2[0][0] = (double)mat[1][1] / deltamat;
        mat2[0][1] = mat[0][1] / -deltamat; //printf("%f\n", mat2[0][1]);
        mat2[1][0] = mat[1][0] / -deltamat;
        mat2[1][1] = (double)mat[0][0] / deltamat;
        printf("%.3lf %.3lf\n", mat2[0][0]*(double)c+mat2[0][1]*(double)f, mat2[1][0]*(double)c+mat2[1][1]*(double)f);
    }
    return 0;
    
}