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) {
  input = input.split(" ");

  var x = parseInt(input[0], 10);
  var k = parseInt(input[1], 10);
  var d = parseInt(input[2], 10);

  var length = x;
  var tmp = x;

  if(x < 0){
    x = -x;
  }

  if (k % 2 == 0) {
    for (var i = 0; i <= k; i += 2) {
      tmp = length - (d * i);
      if (Math.abs(tmp) <= Math.abs(length)) {
        length = tmp;
      } else {
        break;
      }
    }
  } else {
    for (var i = 1; i <= k; i += 2) {
      tmp = length - (d * i);
      if (Math.abs(tmp) <= Math.abs(length)) {
        length = tmp;
      } else {
        break;
      }
    }
  }

  console.log(Math.abs(length));
}

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