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: Runtime Error)

x,k,d = map(int,input().split())
aa = abs(x//d)
if abs(x) < k*d and x < 0:
  ans = x + d * aa
  if (k-aa) % 2 == 0:
    ans = min (abs(ans), abs(ans+d))
  elif (k-aa) % 2 != 0 :
    ans = min (abs(ans-d), abs(ans+d))
elif abs(x) < k*d and x > 0:
  ans = x - d * aa
  if (k -aa) % 2 == 0:
    ans = min (abs(ans), abs(ans-d))
  elif (k -aa) % 2 != 0:
    ans = min (abs(ans-d), abs(ans+d))
elif abs(x) >= k*d and x < 0:
  ans = x + k * d
elif abs(x) >= k*d and x > 0:
  ans = x - k*d

print(ans)