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 Rust (1.42.0) (judge verdict: Wrong Answer)

use proconio::input;

fn main() {
  input! {
    x: isize,
    k: isize,
    d: isize
  }
  
  let result = if x.abs() > (k * d).abs() {
    (x - (k * d)).abs()
  } else {
    let times = if x < d {
      1
    } else {
      (x / d).abs()
    };
    let left = k - times;
    let diff = (x.abs() - (times * d).abs()).abs();
    if diff < x.abs() {
      if left % 2 == 0 {
        diff
      } else {
        std::cmp::min((diff - d).abs(), (diff + d).abs())
      }
    } else {
      if k % 2 == 0 {
        x.abs()
      } else {
        std::cmp::min((x.abs() - d).abs(), (x.abs() + d).abs())
      }
    }
  };
  
  println!("{}", result);
}