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;

// はじめにxは絶対値をとって共通して正にいることにする
// 1. 往復にたどり着く前にKを使い切る
//   - 答え: x - d * k
//   - 制約: x - d * k > 0, d * k でオーバーフロー
//      - x >= dk -> x / d > kで解決！
// 2. 往復する
//   - 往復にたどり着くまで: x / d
//   - その後 -d, +d を繰り返す( k - x / d = k'回 -> 偶数or奇数)

fn main() {
    input! {
        mut x: i64,
        mut k: i64,
        d: i64,
    }

    x = x.abs();

    if (x/d > k) {
        // 往復にたどり着くまでにKを使い切る
        println!("{}", x - d * k);
    } else {
        // 往復し始めるのはe回目
        let e = x / d;
        k -= e;
        x %= d;
        if k % 2 == 1 {
            x = (x - d).abs();
        }
        println!("{}", x);
    }
}