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 Kotlin (1.3.71) (judge verdict: Wrong Answer)

import java.util.*
import kotlin.math.absoluteValue

fun main() {
    val sc = Scanner(System.`in`)
    val x = sc.nextLong()
    val k = sc.nextLong()
    val d = sc.nextLong()
    val result = calc(x, k, d)
    println(result.absoluteValue.toString())
}

fun calc(x: Long, k: Long, d: Long): Long {
    // x < d * k
    var pos = x.absoluteValue

    (1..(k / 2)).forEach {
        pos -= d
        if(pos <= 0) {
            return if(isEven(k - it)) {
                pos
            }
            else {
                pos + d
            }
        }
    }
    return if(isEven(k)) {
        pos
    }
    else {
        pos - d
    }
}

fun isEven(n: Long): Boolean {
    return (n % 2L) == 0L
}