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 (GCC 9.2.1) (judge verdict: Wrong Answer)

#include<stdio.h>
#include<math.h>

int main()
{
  long x, k, d;
  scanf("%ld %ld %ld", &x, &k, &d);
  
  if (x < 0) {
    x *= -1;
  }
  
  double maxMoveDistance = k * d;
  if (x > maxMoveDistance) {
    // 原点に到達できないとき
    printf("%ld\n", x - k * d);
    return 0;
  }
  
  if (x % d == 0) {
    // 原点に移動することがあるとき
    long moveCount = x / d;
    long remainingMoveCount = k - moveCount;
    if (remainingMoveCount % 2 == 0) {
      // 原点にはじめて移動した後の残移動回数が偶数なら原点
      printf("0\n");
      return 0;
    } else {
      // 奇数なら原点から一度移動した座標
      printf("%ld\n", d);
      return 0;
    }
  }
  
  // 原点を跨ぐことができるがちょうど止まることがないとき
  long moveCount = floor(x / d); // 原点を跨ぐ一つ手前の移動回数
  printf("moveCount: %d\n", moveCount);
  long remainingMoveCount = k - moveCount;
  if (remainingMoveCount % 2 != 0) {
    // 残移動回数が奇数なら、そこからもう一回原点方向に移動する(原点を跨ぐ)
    moveCount++;
  }
  x -= d * moveCount;
  
  printf("%ld\n", labs(x));
  
  return 0;
}