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() {
// 文字列
//    var n = readLine()!!.toInt()
//    val x: String = readLine()!!
    // 数値
//    val int =readLine()!!.toInt()
    // 行
//    val list = readLine()!!.split(" ").map (String::toInt )
    var (x,k,d) = readLine()!!.split(" ").map(String::toLong)
//    val xys = List(n) {
////        readLine()!!.split(" ").map(String::toLong)
//        readLine()!!.toDouble()
//    }

    // 0まで移動するための回数
    var zero_step=x / d
    // kがzero_stepよりおおいなら最小値になる可能性あり
    if (zero_step<=k){
        // あまりが答え
        if (k%2==0L){
            print(x)
        }
        print(x%d)
        return
    }else{
        //zero_stepが小さいなら最小値にはならない。掛け算する
        var int_step=zero_step.toInt()
        print(x-(k*d))
        return
    }
//
//    for (i in 0 until k){
//        if (x>=0){
//            x-=d
//        }else{
//            x+=d
//        }
//    }
//    print(Math.abs(x))
}