Problem:
Score : 300 points

Problem Statement

Takahashi, who lives on the number line, is now at coordinate X. He will make exactly K moves of distance D in the positive or negative direction.

More specifically, in one move, he can go from coordinate x to x + D or x - D.

He wants to make K moves so that the absolute value of the coordinate of the destination will be the smallest possible.

Find the minimum possible absolute value of the coordinate of the destination.

Constraints

-10^{15} \leq X \leq 10^{15}

1 \leq K \leq 10^{15}

1 \leq D \leq 10^{15}

All values in input are integers.

Input

Input is given from Standard Input in the following format:

X K D

Output

Print the minimum possible absolute value of the coordinate of the destination.

Sample Input 1

6 2 4

Sample Output 1

2

Takahashi is now at coordinate 6. It is optimal to make the following moves:

Move from coordinate 6 to (6 - 4 =) 2.

Move from coordinate 2 to (2 - 4 =) -2.

Here, the absolute value of the coordinate of the destination is 2, and we cannot make it smaller.

Sample Input 2

7 4 3

Sample Output 2

1

Takahashi is now at coordinate 7. It is optimal to make, for example, the following moves:

Move from coordinate 7 to 4.

Move from coordinate 4 to 7.

Move from coordinate 7 to 4.

Move from coordinate 4 to 1.

Here, the absolute value of the coordinate of the destination is 1, and we cannot make it smaller.

Sample Input 3

10 1 2

Sample Output 3

8

Sample Input 4

1000000000000000 1000000000000000 1000000000000000

Sample Output 4

1000000000000000

The answer can be enormous.

Solution in C (GCC 9.2.1) (judge verdict: Wrong Answer)

#include<stdio.h>
int main(void){
	long int x,k,d,i,j;
	scanf("%ld %ld %ld",&x,&k,&d);
  	for(i=0;i<k;i++){
    	if(x>=0){
        	if(x-k*d >= 0){
            	printf("%ld",x-k*d);
             	return 0;
            }
          	else{
            	j = x/d;
              	x -= j*d;
              	if((k-j)%2 == 1){
                	printf("%ld",(x-d)*(-1));
                  	return 0;
                }
              	else{
                	printf("%ld",x);
                  	return 0;
                }
            }
        }
      	else{
        	if(x+k*d <=0){
            	printf("%ld",(x+k*d)*(-1));
              	return 0;
            }
          	else{
            	j = (-1)*x/d;
              	x += j*d;
              	if((k-j)%2 == 1){
                	printf("%ld",x+d);
                  	return 0;
                }
              	else{
                	printf("%ld",x*(-1));
                  	return 0;
                }
            }
        }
    }
  	return 0;
}