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 PyPy3 (7.3.0) (judge verdict: Wrong Answer)

x, k, d = map(int, input().split())

from math import floor, ceil
def do(x, k, d):
    t1, t2 = floor(x/d), ceil(x/d)
    ogk = k
    ogx = x

    x -= t1*d
    k -= t1
    ans1 = 10**16
    if k%2 == 0:
        ans1 = abs(x)
    else:
        ans1 = min(abs(x + d), abs(x - d))

    x, k = ogx, ogk
    x -= t2*d
    k -= t2
    ans2 = 10**16
    if k%2 == 0:
        ans2 = abs(x)
    else:
        ans2 = min(abs(x + d), abs(x - d))
    
    print(min(ans1, ans2))


if x > 0:
    if x - k*d < 0:
        do(x, k, d)
    else:
        print(x - k * d)

elif x < 0:
    if x + k*d > 0:
        do(abs(x), k, d)
    else:
        print(x + k*d)

else:
    # x == 0
    if k%2:
        print(d)
    else:
        print(0)