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 Fortran (GNU Fortran 9.2.1) (judge verdict: Wrong Answer)

program answer
  implicit none
  integer(8) :: i, x, k, d, count
  read(*,*) X, K, D

if(x==0) then
     if(mod(k,2)==0) then
        write(*,*) D
        stop
     else
        write(*,*) '0'
        stop
     end if
  end if

  if(x<0) then
     if(x+k*d>=0) then
        count=0
        do
           if(x>=0) then
              exit
           end if
           x=x+d
           count=count+1
        end do
        if(mod(count,2)==0) then
           if(mod(k,2)==0) then
              write(*,*) min(abs(x),abs(x-2*d),abs(x+2*d))
              stop
           else
              write(*,*) min(abs(x-d),abs(x+d))
              stop
           end if
        else
           if(mod(k,2)==0) then
              write(*,*) min(abs(x-d),abs(x+d))
              stop
           else
              write(*,*) min(abs(x),abs(x-2*d),abs(x+2*d))
              stop
           end if
        end if
      else
           write(*,*) abs(x+k*d)
           stop
     end if
  else
     if(x-k*d<=0) then
        count=0
        do
           if(x<=0) then
              exit
           end if
           x=x-d
           count=count+1
        end do
        if(mod(count,2)==0) then
           if(mod(k,2)==0) then
              write(*,*) min(abs(x),abs(x+2*d),abs(x-2*d))
              stop
           else
              write(*,*) min(abs(x+d),abs(x-d))
              stop
           end if
        else
           if(mod(k,2)==0) then  
              write(*,*) min(abs(x-d),abs(x+d))
              stop
           else
              write(*,*) min(abs(x),abs(x+2*d),abs(x-2*d))
              stop
           end if
        end if
     else
        write(*,*) x-k*d
        stop
     end if
  end if

  stop
 end program answer