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>

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;
	for(int i=0;i<k;i++){
		if(a==0){//初期座標が正の場合
			if(x>0){
				b=0;
				x = x-d;
			}
			else{
				b=-1;
				x = x+d;
			}
			if(a!=b){//座標値の0を行ったり来たりする事になった場合
				//printf("%lld\n",(k-(i+1))%2);
				if(((k-(i+1))%2)==0){//残り回数が偶数
					if(x<0){
						printf("%lld\n",-x);
						return 0;
					}
					else{
						printf("%lld\n",x);
						return 0;
					}
				}
				else{//残り回数が奇数
					if(a==0){//正から負になった
						if(-x<x+d){//負の座標のほうが絶対値が小さい
							printf("%lld\n",-x);
							return 0;
						}
						else{
							printf("%lld\n",x+d);
							return 0;
						}
					}
					else{
						if(x<x-d){//正の座標のほうが絶対値が小さい
							printf("%lld\n",x);
							return 0;
						}
						else{
							printf("%lld\n",-(x-d));
							return 0;
						}
					}//
				}
			}
		}
		else if(a==-1){//初期座標が負の場合
					if(x>=0){
						b=0;
						x = x-d;
					}
					else{
						b=-1;
						x = x+d;
					}
					if(a!=b){//座標値の0を行ったり来たりする事になった場合
						//printf("%lld\n",(k-(i+1))%2);
						if(((k-(i+1))%2)==0){//残り回数が偶数
							if(x<0){
								printf("%lld\n",-x);
								return 0;
							}
							else{
								printf("%lld\n",x);
								return 0;
							}
						}
						else{//残り回数が奇数
							if(a==0){//正から負になった
								if(-x<x+d){//負の座標のほうが絶対値が小さい
									printf("%lld\n",-x);
									return 0;
								}
								else{
									printf("%lld\n",x+d);
									return 0;
								}
							}
							else{
								if(x<x-d){//正の座標のほうが絶対値が小さい
									printf("%lld\n",x);
									return 0;
								}
								else{
									printf("%lld\n",-(x-d));
									return 0;
								}
							}//
						}
					}
				}

	}
	if(x<0)printf("%lld\n",-x);
	else printf("%lld\n",x);
}