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 Ruby (2.7.1) (judge verdict: Accepted)

# ABC 175 C
# https://atcoder.jp/contests/abc175/tasks/abc175_c

x, k, d = gets.chomp.split(/ /).map(&:to_i)
xp = x.abs

if xp >= d * k
  puts xp - d * k
  exit
end

if k % 2 == 0
  # 0, 2d, 4d, ..., kd 動く
  # i: xp - 2 * i * d > 0 となる最小の i
  # xp > 2 * i * d
  # i < xp / (2 * d)
  i = xp / (2 * d)
  puts [xp - 2 * i * d, xp - 2 * (i + 1)*d].map(&:abs).min
else
  # d, 3d, 5d, ..., kd 動く
  # i: xp - (2 * i + 1) * d > 0 となる最小の i
  # xp > (2 * i + 1) * d
  # xp - d > 2 * i * d
  # i < (xp - d) / (2 * d)
  i = (xp - d) / (2 * d)
  puts [xp - (2 * i + 1)*d, xp - (2 * i + 3)*d].map(&:abs).min
end