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 Scala (2.13.1) (judge verdict: Wrong Answer)

object Main extends App {
  val xkdInput = scala.io.StdIn.readLine()
  val xkdArray = xkdInput.split(" ").map(_.toLong).toVector

  val x = xkdArray(0)
  val k = xkdArray(1)
  val d = xkdArray(2)

  var tempK = k
  var tempMin = x

  if (tempMin < 0) {
    var kMinus = (Math.abs(x) / d)
    if (kMinus > k) {
      kMinus = k
      tempK = 0
    } else {
      tempK = kMinus
    }
    tempMin = x + kMinus * d

  } else if(tempMin > 0) {
    var kMinus = (Math.abs(x) / d)
    if (kMinus > k) {
      kMinus = k
      tempK = 0
    } else {
      tempK = kMinus
    }
    tempMin = x - kMinus * d
  }

  if(tempK == 0) {
    println(Math.abs(tempMin))
  } else if (tempK % 2 == 0) {
    println(Math.abs(tempMin))
  } else {
    if (tempMin < 0) {
      println(Math.abs(tempMin + d))
    } else {
      println(Math.abs(tempMin - d))
    }
  }
}