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 <stdlib.h>
#include <string.h>
#include <math.h>
#include <ctype.h>
#include <inttypes.h>

//int compare_int(const void *a, const void *b){//qsort 整数比較<
//	return *(int*)a - *(int*)b;
//}
//int compare_int2(const void *a, const void *b){//qsort 整数比較>
//	return *(int*)b - *(int*)a;
//}
//
int main(void){
	double x;
	long long int a,k,d;
	long long int b;

	scanf("%lf %lld %lld",&x,&k,&d);
	a=x/d;
	if(a>k){
		if(x>0){
			x=x-k*d;
		}else{
			x=x+k*d;
		}
		b=(long long int)x;
		if(b<0){
			b=-b;
		}
		printf("%lld",b);
		return 0;
	}
	x=x-a*d;
	k=k-a;
	if((long long int)k%2==1){
		if(abs(x+d)>abs(x-d)){
			b=(long long int)(x-d);
			if(b<0){
				b=-b;
			}
			printf("%lld",b);
		}else{
			b=(long long int)(x+d);
			if(b<0){
				b=-b;
			}
			printf("%lld",b);
		}
	}else{
		b=(long long int)x;
		if(b<0){
			b=-b;
		}
		printf("%lld",b);
	}

	return 0;
}