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>
#include <math.h>

int main(void)
{
  long long int a, b, c, x, k, d, m = 0, result = 0;
  
  scanf("%lld %lld %lld", &x, &k, &d);

  
  if( x < 0 ){
      m = 1;
      x *= -1;
  } else {
      if( x ==  0 ){
          if( k % 2 == 1 ){
              printf("%lld\n", d);
              return 0;
          } else {
              printf("0\n");
              return 0;
          }
      }
  }
  
  
  if( x <= d ){
    b = x % d; 
    a = d - b;
    c = 0;
  } else {
      a = x % d; 
    b = d - a;
    c = (x-a) / d;
  }
  //printf("%lld %lld %lld\n", a, b, c);
  
  if( x >= k*d ){
      result = x - k*d;
  } else if( k - c % 2 != 1 ){
          result = a;
      } else {
          result = b;
      }
  
  printf("%lld\n", result);
  
  return 0;
}