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::input;

fn main() {
    input! {
        mut x: i128,
        k: i128,
        d: i128,
    }

    solve(x, k, d);
}

fn solve(mut x: i128, k: i128, d: i128) {
    // x は正として良い
    x = x.abs();

    // x が遠すぎてどれだけ引いても原点までたどり着かない場合
    if x >= k * d {
        println!("{}", x - k * d);
        return;
    }

    // 適当な回数で原点までたどり着く場合
    if x < k * d {
        // x が正である最小の値を z とする
        // z にたどり着くために動かした回数を y とする
        let y = x / d;
        let z = x - y * d;

        // 移動の残回数が偶数回の場合は、点 z が最小になるしかない
        if (k - y) % 2 == 0 {
            println!("{}", z);
            return;
        }
        // 移動の回数が奇数回の場合は、必ず移動しないといけないので、原点方向に動く
        else {
            println!("{}", (z - d).abs());
            return;
        }
    }
}