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>
#define INF 2000000000
#define MOD 1000000007
#define EPS (1e-10)

using namespace std;

int main(int argc, char *argv[]) {

	long X,K,D; cin >> X >> K >> D;

	long ans = 0;
	if (abs(X) >= K * D ) {
		ans = abs(X) - (K * D);
	}else {
		if (abs(X) % D == 0) {
			K -= abs(X) / D;		
			if (K % 2 == 0) {
				ans = 0;	
			}else{
				ans = D;
			}
		}else {
			K -= abs(X) / D;
			if (K % 2 == 0) {
				ans = abs(X) - (abs(X) / D) * D;	
			}else {
				ans = abs(abs(X) - (abs(X) / D) * D - D);	
			}
		}
		//long temae_made = (abs(X) / D);
		//K -= temae_made;
		//ans = abs(X) - (temae_made * D);		
		//if (K % 2 == 1) {
		//	ans = abs(D - ans);
		//}
	}
	cout << ans << endl;

		

	return 0;
}