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: Accepted)

#include<stdio.h>
#include<stdlib.h>
#include<math.h>
#define MAX 1000001
#define abs(y) ((y>0) ? (y) : (-(y)))

//int solve(int data[],int size);
typedef long long int int64;

int main()
{
    int64 x,k,d;
    int i,j;
    int a=7;
	scanf("%lld %lld %lld",&x,&k,&d);

	//遠い場合は近づく

	if(x<0)x= x * (-1);

	if(x>0)
    {
	    if(x/d<=k)//間に合う
        {
	        int64 cand1 = x%d;
	        int64 cand2 = abs(((x%d)-d));
	        if((k-(x/d))%2==0)printf("%lld",cand1);
	        else printf("%lld",cand2);
        }
	    else//間に合わない
        {
	        printf("%lld",x-(k*d));
        }
    }
	else if(x==0)
    {
	    if(k%2==0)printf("0");
	    else printf("%lld",abs(d));

    }
	//printf("%d",i);
}