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>


int main(){
	long i,j,k;
	long X,K,D;
	scanf("%ld%ld%ld",&X,&K,&D);
	
	long S=labs(X)/D;
	if(S<K) {
		if(X>0)X -= D*S;
		else X+=D*S;
	}
	else {
		if(X>0)
		X -=D*K;
		else X+=D*K;
		S=K;
	}
	for(i=S+1;i<K;i++){
		if(X>0){
			X-=D;
			//("X:%ld,",X);
			if(X<0)break;
		}
		else {
			X+=D;//printf("X:%ld,",X);
			if(X>0)break;
		}
	}
	//printf("i:%ld,K:%ld\n",i,K);
	if(i!=K)
	if((K-i-1)%2!=0){
		if(X>0)X-=D;
		else X+=D;
	//	printf("aX:%ld,",X);
	}
	printf("%ld\n",labs(X));
	return 0;
}