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;
#[allow(unused_imports)]
use proconio::marker::{Bytes, Chars};
#[allow(unused_imports)]
use std::cmp::{min, max};

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

	let xabs = x.abs();
	let s = xabs / d;
	if s >= k {
		if 0 < x {
			println!("{}", x - k * d);
			return;
		}
		else {
			println!("{}", (x + k * d).abs());
			return;
		}
	}

	let tx;
	let tk;
	if 0 < x {
		tx = x - s * d;
		tk = k - s;
		if tk % 2 == 0 {
			println!("{}", tx);
			return;
		}
		else {
			println!("{}", (tx - d).abs());
			return;
		}
	}
	else {
		tx = x + s * d;
		tk = k - s;
		if tk % 2 == 0 {
			println!("{}", tx.abs());
			return;
		}
		else {
			println!("{}", tx + d);
			return;
		}
	}
}