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)]
use proconio::input;
use std::cmp::min;

fn main() {
    input! {
        mut X: i64,
        mut K: i64,
        D: i64,
    }

    // 2回消費して空撃ちすることが可能。
    // 戦略としては、符号を維持するように移動 → 微調整 → 空打ちで消費か?

    // 絶対値を見るので、Xは正としてよい。
    if X < 0 {
        X = -X;
    }

    // let mut ans = std::i64::MAX;

    let cnt = min(X / D, K);
    K -= cnt;
    X -= D * cnt;

    if K == 0 {
        println!("{}", X);
        return;
    }

    // ここまでで、[0, D]の区間にいるはず
    // 奇数だったら1回、偶数だったら0回やっておしまい
    if K % 2 == 0 {
        println!("{}", X);
        return;
    } else {
        println!("{}", (X - D).abs());
        return;
    }
}