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: i64,
        k: i64,
        d: i64,
    }

    solve(x, k, d);
}

fn solve(x: i64, k: i64, d: i64) {
    // スタート位置は必ず正として良い
    let mut ans = x.abs();
    for i in 0..k {
        // まず移動する
        ans -= d;

        // 点の位置が負になる場合
        if ans <= 0 {
            // 今いる位置 or 負の値が答え

            // 残りの移動回数が奇数回の場合、
            // 行き来するしかない
            if (k - i) % 2 == 0 {
                println!("{}", ans + d);
                return;
            }
            // 残りの移動回数が偶数回の場合
            else {
                println!("{}", ans.abs());
                return;
            }
        } else {
            ans -= d;
        }
    }

    println!("{}", ans.abs());
}