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 <math.h>
#include <stdlib.h>

int main(void){
    long long x,l,d,ans,i=0,k=0;
    scanf("%lld %lld %lld", &x,&l,&d);

    ans = x;
    if(x>0){
        if(x - l*d > 0){
            printf("%lld\n", x - l*d);
            return 0;
        }
        while(ans>=0){
            k = ans;
            ans -= d;
            i++;
            if(i==l) {
                break;
            }

        }
        if((i-l)%2 != 0){
            printf("%lld\n", llabs(k));
            return 0;
        }else if((i-l)%2 == 0){
            printf("%lld\n", llabs(ans));
            return 0;
        }
    }else{
        while(ans<=0){
            k = ans;
            ans += d;
            i++;
            if(i==l) {
                break;
            }
        }

        if((i-l)%2 != 0){
            printf("%lld\n", llabs(k));
            return 0;
        }else if((i-l)%2 == 0){
            printf("%lld\n", llabs(ans));
            return 0;
        }
    }
    return 0;
}