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: Accepted)

using System;
using System.Linq;

class program
{
    static void Main()
    {
            var line = Console.ReadLine().Split(' ').Select(x => long.Parse(x)).ToArray();

            long X = line[0]; // 現在地
            var K = line[1]; // 移動回数
            var D = line[2]; // 移動距離

            long ans = X; // K回後の最小の絶対値

            // moveTimesToRoundTrip: 往復移動をするまでに必要な移動回数
            // Xがマイナスでも答えは変わらないため絶対値で計算
            // 切り捨ててもKは整数であるため同じ値になる
            var moveTimesToRoundTrip = Math.Abs(X) / D;

            if (moveTimesToRoundTrip >= K)
            {
                // 往復移動をしない場合
                // 現在地から移動回数*移動距離を足し引きして求める
                if (X >= 0)
                {
                    ans = X - K * D;
                }
                else
                {
                    ans = X + K * D;
                }
            }
            else
            {
                // 往復移動をする場合

                // distanceBeforeRoundTrip: 往復移動する前の距離
                var distanceBeforeRoundTrip = X % D;

                // roundTripTimes: 往復回数
                // 原点近くにたどり着くまでにmoveTimesToRoundTripかかる
                // 残りの回数roundTripTimesは原点をまたいで-D, +Dを繰り返す
                var roundTripTimes = K - moveTimesToRoundTrip;

                if (roundTripTimes % 2 == 0)
                {
                    // 偶数回移動するなら
                    // 往復移動しても往復移動する前の距離が答え
                    ans = distanceBeforeRoundTrip;
                }
                else
                {
                    // 奇数回移動するなら
                    // 往復移動する前の距離が正なら移動距離Dを引いて
                    // 往復移動する前の距離が負なら移動距離Dを足す
                    if (distanceBeforeRoundTrip > 0)
                    {
                        ans = distanceBeforeRoundTrip - D;
                    }
                    else
                    {
                        ans = distanceBeforeRoundTrip + D;
                    }
                }
            }

            Console.WriteLine(Math.Abs(ans));
                
    }
}