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 JavaScript (Node.js 12.16.1) (judge verdict: Wrong Answer)

function Main(input) {
  const row = input.split(" ");
    	let X = parseInt(row[0], 10);
    	let K = parseInt(row[1], 10);
    	const D = parseInt(row[2], 10);
    let absX = Math.abs(X);
    //let a = absX / D;//往復にたどり着いた座標
    let ans; //答え

    if((absX /D) >= K) {
       ans = absX - D * K;//往復地点までたどり着かな勝った時の計算 x-dk >=0→x/d>=k

    } else {
      //0近辺の往復処理を書く.K2 =K -(X/d)で往復地点までかかった回数を減らして、残りの回数分往復する
       K2 = K - absX / D;//Kからx/d回引いた回数が「往復を繰り返す回数」
       absX = Math.abs(absX - (absX / D) * D);//x/D回-Dした後の往復開始座標
      if(K2 % 2 === 1) {
    　ans = absX - D;
     } else {
         ans = absX;
      }
    } console.log(Math.abs(ans));
}

Main(require("fs").readFileSync("/dev/stdin", "utf8"));