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 whiteread::parse_line;
use std::process;

fn main() {
    let (x, k, d): (i64, i64, i64) = parse_line().unwrap();
    let mut p: i64 = x;
    let mut i: i64 = 0;
    if x == d && k % 2 == 0 {
        println!("{:?}", x);
        process::exit(0x0100);
    }

    let mut v: Vec<i64> = vec![p,p-1,p-1];

    while i < k {
        if p == 0 && i % 2 == 0 {
            break;
        } else if p.abs() <= d {
            break;
        } else if v[0] == v[2] {
            break;
        } else {
            if p >= 0 {
                p = p - d;
            } else {
                p = p + d;
            }
            v[0] = v[1];
            v[1] = v[2];
            v[2] = p;
        }
        i = i + 1;
    }

    println!("{:?}", p.abs());
}