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 kotlin.math.abs

fun main(args: Array<String>) {
    
    val (x, k, d) = readLine()!!.split(" ").map { it.toLong() }
    

    //val (x, k, d) = Triple<Long, Long, Long>(7, 1, 10)

    val needBestTimes = if (x > d) abs(x) / d else 1L
    val bestPos = if (x > d) abs(x) % d else d - abs(x)

    val answer = when {
        needBestTimes > k -> { // OK
            abs(x) - k * d
        }
        (k - needBestTimes) % 2 == 0L -> {
            bestPos
        }
        (k - needBestTimes) % 2 == 1L -> {
            bestPos + d
        }

        else -> {
            0L
        }
    }

    println("$answer")
}