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 = list(map(int,input().split()))
count = K
ans = abs(X)
x = X
L = K//2
mx = X//D

if x>0:
    if mx<K:
        x = X-mx*D
        count -= mx
    else:
        x = X-(K-1)*D
        count -= K-1
    while x>0 and count>0:
        x -= D
        count -= 1
        if abs(x)<ans and count%2==0:
            ans = abs(x)
elif x<0:
    if mx<K:
        x = X+mx*D
        count -= mx
    else:
        x = X+(K-1)*D
        count -= K-1
    while x<0 and count>0:
        x += D
        count -= 1
        if abs(x)<ans and count%2==0:
            ans = abs(x)

print(ans)