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 long int x,d,ans,k,i;
  int a=1,b=-1;ans=1,
  
  
  scanf("%lld %lld %lld",&x,&k,&d);
 
  if(x<0){x=-x;}
  if(k%2==0){b=0;}
  if(x-k*d>=0){ans=x-k*d;}
  
  else{
  for(i=b;i<=k;i+=2){
    ans=x-i*d;
    if(ans<=0)break;
  }
  
 
  
  if(ans+ans+2*d<=0){ans=ans+2*d;}
  if(ans<0) ans=-ans;
  }
printf("%lld",ans);
  return 0;
}