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)

use proconio::*;
#[allow(unused_imports)]
use proconio::marker::*;

#[allow(unused_macros)]
macro_rules! echo {
    ($ x : expr ) => {
        println!("{}", $x);
    };
}

#[fastout]
#[allow(non_snake_case)]
fn main() {
    input! {
        X: i128,
        K: i128,
        D: i128,
    }
    if X.abs() >= K * D {
        echo!((X.abs() - K * D).abs());
    } else {
        let toutatu_count = X.abs() / D;
        let remain_K = K - toutatu_count;
        let anss = vec![(X.abs() - D * toutatu_count).abs(), (X.abs() - D * (toutatu_count + 1)).abs()];
        echo!(anss[(remain_K % 2) as usize]);
    }
}