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 <iostream>
#include <string>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>

using namespace std;
int main()
{
    // 文字列の入力
    long long X, K, D;
	cin >> X >> K >> D;
	
	X = abs(X);
	long long result;
	if(abs(X) / D >= K) {
		// 折り返しなし
		result = (X > 0) ? (X - K * D) : (-X + K * D);
		printf("%d\n", result);
		return 0;
	}
	
	long long div = X / D;
	long long sur = X % D;
	
	if((K - div) % 2 == 0) {
		result = sur;
	} else {
		result = (sur > 0) ? (sur - D) : (sur + D);
	}
	
	printf("%lld\n", abs(result));
	
    return 0;
}