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>
int main(void)
{
    long long int n,x,k,d,i;
    scanf("%lld %lld %lld",&x,&k,&d);
    if(x > 0)
    {
        if(x - d * k >= 0)
        {
            printf("%lld\n",x - d * k);
            return 0;
        }
        if(x - d * k < 0)
        {
            int a;
            a = x / d;
            x = x - d * a;
            if((k - a) % 2 == 0)
            {
                printf("%lld\n",x);
                return 0;
            }
            else
            {
                printf("%lld\n",d - x);
                return 0;
            }
        }
    }
    if(x == 0)
    {
        if(k % 2 == 0)
        {
            printf("0\n");
            return 0;
        }
        else
        {
            printf("%lld\n",d);
            return 0;
        }
    }
    if(x < 0)
    {
        x = -1 * x;
        if(x - d * k >= 0)
        {
            printf("%lld\n",x - d * k);
            return 0;
        }
        if(x - d * k < 0)
        {
            int a;
            a = x / d;
            x = x - d * a;
            if((k - a )% 2 == 0)
            {
                printf("%lld\n",x);
                return 0;
            }
            else
            {
                printf("%lld\n",d - x);
                return 0;
            }
        }
    }
    return 0;
}