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: Time Limit Exceeded)

x,k,d = gets.chomp.split(' ').map(&:to_i)
flag = ""
if x > 0
  if x - (d * k) >= 0 #kが大きく、dが小さい時対策
    puts x - (d * k)
    flag = "ok"
  end
elsif x < 0
  if x + (d * k) <= 0 #kが大きく、dが小さい時対策
    puts (x + (d * k)).abs
    flag = "ok"
  end
end
i = 0
while i < k
  if (x - d).abs <= (x + d).abs #dを引いたほうが近づくとき、
    x = x - d #それを改めてxとする
    if x <= 0 && x + d > 0#符号を跨いでいたなら、
      cand = [x.abs, x+d] #その前後が候補となる。配列の第一要素が、i回目の操作の結果であることに注意する。
      # puts "break!+→-"
      # puts cand
      # puts "===="
      break
    end
  else
    x = x + d
    if x >= 0 && x - d < 0
      cand = [x,(x-d).abs]
      # puts "break!-→+"
      # puts cand
      # puts "===="
      break
    end
  end
  i += 1
end
if flag != "ok"
  if i == k
    puts x.abs
  else
    if (k - i) % 2 == 1
      # puts "sign"
      puts cand[0]
    else
      # puts "sing"
      puts cand[1]
    end
  end
end