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.*

fun main() {
    val list = readLine()!!.split(" ")
    val x = list[0].toLong()
    val k = list[1].toLong()
    val d = list[2].toLong()
    val minimum: Double = x.toDouble()/1000000.0- k * d.toDouble()/1000000.0
    val maximum: Double = x.toDouble()/1000000.0 + k * d.toDouble()/1000000.0
    if ((x * minimum > 0 && x * maximum > 0)) {
        if (x < 0) {
            println(x + k * d)
        } else if (x > 0) {
            println(x - k * d)
        }
    } else if (x * minimum < 0) {
        val howMany: Long = x / d
        val nam: Long = k - howMany
        if (nam % 2 == 1L) {
            println(min(abs(x - (howMany - 1) * d), abs(x - (howMany + 1) * d)))
        } else {
            println(x - howMany * d)
        }

    } else if (x * maximum < 0) {
        val howMany: Long = (-x) / d
        val nam: Long = k - howMany
        if (nam % 2 == 1L) {
            println(min(abs(x + (howMany - 1) * d), abs(x + (howMany + 1) * d)))
        } else {
            println(x + howMany * d)
        }


    }
}