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: Accepted)

#[allow(unused_imports)]
use {
   proconio::{fastout, input, marker::*},
   std::cmp::*,
   std::collections::*,
};

#[fastout]
fn main() {
   input! {
      mut x: i64,
      mut k: usize,
      mut d: i64,
   }
   if k % 2 == 1 {
      if x > 0 {
         x -= d;
      } else {
         x += d;
      }
      k -= 1;
   }

   if x < 0 {
      x *= -1;
   }

   let m = 2 * (x / 2 / d) as usize;
   if m >= k {
      println!("{}", x - k as i64 * d);
   } else {
      println!(
         "{}",
         min(
            (x - (m + 2) as i64 * d).abs(),
            (x - m as i64 * d).abs()
         )
      );
   }
}