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: Time Limit Exceeded)

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

int main(void) {
	long long int x,k,d;
	scanf("%lld%lld%lld",&x,&k,&d);
	int a=0,b=0;
	if(x<0)a=-1,b=-1;
	else if(x==0){//初期座標が0
		if(k%2==0){
			printf("0\n");
			return 0;
		}
		else{
			printf("%lld\n",x+d);
			return 0;
		}
	}
	if(a==0){//初期座標が正の値
		while(0<x&&0<k){
			x -= d;
			k--;
		}
		if(k==0){
			if(x>=0){
				printf("%lld\n",x);
				return 0;
			}
			else{
				printf("%lld\n",-x);
				return 0;
			}
		}
		else if(k%2==0){//0を跨いだときの残り回数が偶数
			printf("%lld\n",-x);
			return 0;
		}
		else{//奇数
			if(-(x-d)<x+d){
				printf("%lld\n",-(x-d));
				return 0;
			}
			else{
				printf("%lld\n",x+d);
				return 0;
			}
		}
	}
	else{//初期値が負の値
		while(x<0&&0<k){
			x += d;
			k--;
		}
		if(k==0){
			if(x>=0){
				printf("%lld\n",x);
				return 0;
			}
			else{
				printf("%lld\n",-x);
				return 0;
			}
		}
		else if(k%2==0){//0を跨いだときの残り回数が偶数
			printf("%lld\n",x);
			return 0;
		}
		else{//奇数
			if(x+d<-(x-d)){
				printf("%lld\n",x+d);
				return 0;
			}
			else{
				printf("%lld\n",-(x-d));
				return 0;
			}
		}
	}
	return 0;
}