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: Accepted)

//let input = "100 22 5"; // x k d 初期座標　回数　一度に動ける

function main(input){
  input = input.trim();
  let list = input.split(" ");
  let x = Math.abs(parseInt(list[0]));
  let k = parseInt(list[1]);
  let d = parseInt(list[2]);
  let kaisuu = Math.floor(x/d); // ０を過ぎずに最も接近できる回数
  if(kaisuu>k){ // kを超えてしまったらkaisuuはk
    kaisuu=k;
  }
  let nokoriKaisuu = k - kaisuu;
  if(nokoriKaisuu%2===0){ // 残り回数が偶数のとき
    // 何もしない
  }else if(nokoriKaisuu%2===1){ // 残り回数が奇数のとき
    if(0 < x - d*kaisuu){ // kaisuu回移動したとき、xが正
      kaisuu+=1;
    }else{ // xが０か負
      kaisuu-=1;
    }
  }
  let answer = Math.abs(x - d*kaisuu);
  console.log(answer);
}
//main(input);
main(require("fs").readFileSync("/dev/stdin", "utf8"));