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)

/* ABC175 C */

#include<stdio.h>
#include<stdlib.h>

int main(void){
    long long x, k, d, x1, x2, y1, y2, q;

    scanf("%lld %lld %lld", &x, &k, &d);

    q = x/d;
    if(q<k){    //0に近づいても試行回数が余るとき
        k -= q;
        x %= d;
    }else{      //0に近づけないまま試行が終わるとき
        x -= k*d;
        k = 0;
    }

    while(0<k){
        x1 = x+d;
        x2 = x-d;
        y2 = y1;
        y1 = x;
        k--;
        if(llabs(x1)>llabs(x2)) x = x2;
        else x = x1;

        //ループするときの処理
        if(x==y2){
            k %= 2;
        }
    }
    printf("%lld\n", llabs(x));
}