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)

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

    val t = Math.abs(x) / d
    val tr = Math.abs(x) % d

    if (t >= k) {
        println(Math.abs(x) - (d * k))

    } else {
        if (tr == 0L) {
            if ((k - t) % 2L == 0L) {
                println(0)
            } else {
                println(d)
            }
        } else {
            if (tr < d - tr) {
                if ((k - t) % 2L == 0L) {
                    println(tr)
                } else {
                    println(d - tr)
                }
            } else {
                if ((k - (t + 1)) % 2L == 0L) {
                    println(d-tr)
                } else {
                    println(d)
                }
            }
        }
    }
}