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 (Clang 10.0.0) (judge verdict: Wrong Answer)

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

long long int absolute(long long int a) {
	if(a > 0) {
		return a;
	} else {
		return -a;
	}
}

int main(void) {
	int i;
	long long int X, K, D;
	char buf[256], *ch;

	fgets(buf, sizeof(buf), stdin);
	ch = strtok(buf, " ");
	X = atoll(ch);
	ch = strtok(NULL, " ");
	K = atoll(ch);
	ch = strtok(NULL, " ");
	D = atoll(ch);

	if(K < (X / D)) {
		printf("%lld\n", (X - (K * D)));
	} else if((K - (X / D)) % 2 == 0) {
		printf("%lld\n", (X % D));
	} else {
		printf("%lld\n", absolute(D - (X % D)));
	}

	return 0;
}