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(unused_imports)]

use std::cmp::{max, min, Ordering};
use std::collections::*;
use std::{i32, i64, i128, u32, u64, u128, f64};
use std::io::{stdout, Write, BufWriter};
use std::mem::{swap};
use proconio::marker::{Isize1, Usize1, Bytes, Chars};

#[proconio::fastout]
fn main() {
    proconio::input!(
        x: i64,
        k: i64,
        d: i64,
    );
    let x = x.abs();
    let (q, r) = (x / d, x % d);
    if k <= q { println!("{}", x - k*d); }
    else {
        if (k - q) % 2 == 0 { println!("{}", r); }
        else { println!("{}", d - r); }
    }
}