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 <bits/stdc++.h>
using namespace std;

int main(){
	double X, K, D;
	cin >> X >> K >> D;
	if(X>0){
		double num = floor(X/D);
		if(num > K){
			printf("%.0lf\n",abs(X-K*D));
			return 0;
		}
		double buf1 = abs(X-num*D);
		double buf2 = abs(X-(num+1)*D);
		if(buf1 < buf2){
			if(floor((K-num)/2)==(K-num)/2){
				printf("%.0lf\n", buf1);
				return 0;
			}else{
				printf("%.0lf\n", buf2);
				return 0;
			}
		}else{
			if(floor((K-num-1)/2)==(K-num-1)/2){
				printf("%.0lf\n", buf2);
				return 0;
			}else{
				printf("%.0lf\n", buf1);
				return 0;
			}
		}
	}else if(X<0){
		double num = floor(abs(X)/D);
		if(num > K){
			printf("%.0lf/n",abs(X+K*D));
			return 0;
		}
		double buf1 = abs(X+num*D);
		double buf2 = abs(X+(num+1)*D);
		if(buf1 < buf2){
			if(floor((K-num)/2)==(K-num)/2){
				printf("%.0lf\n", buf1);
				return 0;
			}else{
				printf("%.0lf\n", buf2);
				return 0;
			}
		}else{
			if(floor((K-num-1)/2)==(K-num-1)/2){
				printf("%.0lf\n", buf2);
				return 0;
			}else{
				printf("%.0lf\n", buf1);
				return 0;
			}
		}
	}else if(X==0){
		if(floor(D/2)==D/2){
			cout << 0 << endl;
			return 0;
		}else{
			printf("%.0lf\n",D);
			return 0;
		}
	}

}