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 C (GCC 9.2.1) (judge verdict: Wrong Answer)


#include<stdio.h>
#include<string.h>

int X,K,D;

int main(void){
    scanf("%d""%d""%d", &X,&K,&D);
    int ans=0;
    int n;
    int min;

    if (X<0)X=-X;
    
    n=X/D;
    int num1=K-n;
    int num2=K-(n+1);



    if(n>=K) ans =X-K*D;
    else if (X==D*n){
        ans=X;
    }
    else if (X-D*n<=K/2){
        if (num1%2==0) ans=X-D*n;
        else ans= D*(n+1)-X;

    }
    else if (X-D*n<K/2){
        if (num2==0) ans= D*(n+1)-X;
        else ans=X-D*n;
    }
    
    
     printf("%d\n",ans);
  
return 0;
}