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 Python (3.8.2) (judge verdict: Wrong Answer)

x,k,d=map(int,input().split())
sa=abs(x)//d
ans=x
pbk,mbk=0,0
if sa>=k:
  ans=abs(x)-d*k
else:
  ans=abs(x)-d*sa
  for i in range(k-sa):
    if abs(ans-d)<abs(ans+d):
      ans=abs(ans-d)
    elif abs(ans-d)>abs(ans+d):
      ans=abs(ans+d)
    else:
      if (k-i)%2==1:
        ans=d
      break
    if abs(ans-d)>abs(ans+d):
      if pbk==abs(ans-d) and mbk==abs(ans+d):
        if (k-i)%2==1:
          ans=mbk
          break
      pbk=abs(ans-d)
      mbk=abs(ans+d)
    else:
      if pbk==abs(ans+d) and mbk==abs(ans-d):
        if (k-i)%2==1:
          ans=mbk
          break
      pbk=abs(ans+d)
      mbk=abs(ans-d)
print(ans)