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 java.lang.Long.min
import java.math.BigInteger

private fun getLongs(): LongArray {
    return readLine()!!.split(" ").map {it.toLong()}.toLongArray()
}


private fun absL(l:Long):Long {
    return if (l<0) {-l} else {l}
}

private fun solver(x:Long,k:Long,d:Long):Long{
    if (k % 2 == 0L) {
        return evensolve(x,k,d)
    } else {
        return oddsolve(x,k,d)
    }
}
private fun evensolve(x:Long,k:Long,d:Long):Long {
    val xx = absL(x)
    if (condBigInt(xx,k,d)) {
        return xx - k*d
    }
    // 原点を通り過ることができる。
    // 実質的に2dグリッドで動ける。
    // 原点を通り過ぎない。
    val a = xx % (2 * d)
    // 通り過ぎる場合。
    val b = (2 * d) - a
    return min(a,b)
}

private fun oddsolve(x:Long,k:Long,d:Long):Long {
    // 最初の動く方向を右左どちらかに固定した場合偶数に帰着。
    return min(evensolve(x+d,k-1,d),evensolve(x-d,k-1,d))
}

fun main(){
    print(Long.MAX_VALUE)
    val v = getLongs()
    print(solver(v[0],v[1],v[2]))
}

private fun condBigInt(x:Long,k:Long,d:Long):Boolean {
    return x.toBigInteger() >= k.toBigInteger() * k.toBigInteger()
}