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
import kotlin.math.min

fun main(){
    val (x, k, d) = readLine()!!.split(" ").map { it.toLong() }
    if (x >= 0){
        val saitan = x / d
        if (k <= saitan){
            if (k <= 100000000){
                var ans = x
                for (i in 0 until k){
                    ans -= d
                }
                println(ans)
            } else {
                println(abs(x - k * d))
            }
        } else {
            if ((k - saitan) % 2 == 0L){
                println(abs(x % d))
            } else {
                println(abs(x % d - d))
            }
        }
    } else {
        val useX = x * -1
        val saitan = useX / d
        if (k <= saitan){
            if (k <= 100000000){
                var ans = x
                for (i in 0 until k){
                    ans += d
                }
                println(ans)
            } else {
                println(abs((k * d - x)))
            }
        } else {
            if ((k - saitan) % 2 == 0L){
                println(abs(x % d))
            } else {
                println(abs(x % d + d))
            }
        }
    }
}