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: Time Limit Exceeded)

use std::io;

fn main() {
    let mut S = String::new();
    io::stdin().read_line(&mut S);
    let vec :Vec<&str> = S.trim().split_whitespace().collect();
    let mut x :i64 = vec[0].parse().unwrap();
    let mut k :usize = vec[1].parse().unwrap();
    let d :i64 = vec[2].parse().unwrap();
    if (x < 0){
        while (x < 0 && k > 0){
            x += d;
            k -= 1;
        }
        if(k % 2 == 0){
            println!("{}",x.abs());
        }else{
            println!("{}",(x - d).abs());
        }
    }else{
        while (x > 0 && k > 0){
            x -= d;
            k -= 1;
        }
        if(k % 2 == 0){
            println!("{}",x.abs());
        }else{
            println!("{}",(x + d).abs());
        }
    }
}