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 C# (.NET Core 3.1.201) (judge verdict: Wrong Answer)

using System;
using System.Linq;
//x<0の時絶対値をとって処理を共通化する
//場合分け
//(1)原点周りでの往復にたどり着く前に終わる 答え= x-d*k
//条件=x-d*k>=0
//ただしd*kはオーバーフローする->x/d>=kとすればok
//(2)原点周りで終わる
//x/d回は-d,そこから(残り操作回数k-x/d=k')は-d,+d,-d,+dしたがってk'の偶数奇数が知りたい

class Program
{
    static void Main()
    {
        long[] In = Console.ReadLine().Split(' ').Select(x => long.Parse(x)).ToArray();
        long x = In[0];
        long k = In[1];
        long d = In[2];
        if (x < 0) { x *= -1; d *= -1; }
        long c = x / d;
        //x-d*k>=0の時
        if (c >= k) Console.WriteLine(x-d*k);
        else
        {
            //繰り返しの前の位置
            long p = x % d;
            //残数
            long t = k - c;
            if (t % 2 == 0)
            {
                Console.WriteLine(Math.Abs(p));
            }
            else
            {
                Console.WriteLine(Math.Abs(p - d));
            }
        }
    }
}