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(
  non_snake_case,
  unused_variables,
  unused_assignments,
  unused_mut,
  unused_imports,
  dead_code
)]
use proconio::{input, marker::*};
use std::cmp::*;
use std::collections::*;
//use std::num;
//use petgraph::unionfind::UnionFind;
//use permutohedron::LexicalPermutation as _;

//#[fastout()]
fn main() {
  input! {
      X:i64,K:i64,D:i64
  }
  let a = X.abs() / D;
  // a or a+1回移動した時点で原点に一番近づく。
  if K <= a {
    let mut ans = X.abs() - D * K;
    println!("{}", ans.abs());
    return;
  }
  // a+1回以上の移動が可能。
  // 後K-a回移動する。
  let b = if (K - a) % 2 == 0 { a } else { a + 1 };
  let mut ans = X.abs() - D * b;
  println!("{}", ans.abs());
}