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() {
    abc175c()
}

fun abc175c() {
    val (x, k, d) = readLine()!!.split(" ").map { it.toLong() }
    var temp = x / d
    if (temp >= k) {
        println(x - (d * k))
        return
    }
    var remain = k - temp
    var t = x - temp * d
    if (t > abs(t - d)) {
        t = abs(t - d)
        remain--
    }
    if (remain == 0L) {
        println(abs(t))
        return
    }
    if (remain % 2 == 0L) {
        // この座標が答え
        println(abs(t))
    } else {
        // 0をまたいだプラス方向が答え
        println(abs(t - d))
    }
}