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 <math.h>
#include <stdio.h>
typedef long long ll;
ll ab(ll x) {
	return x > 0 ? x : -x;
}
ll min(ll x, ll y) {
	return x > y ? y : x;
}
ll max(ll x, ll y) {
	return x < y ? y : x;
}
int main() {
	ll x, k, d;
	scanf("%lld%lld%lld", &x, &k, &d);
	if (log(ab(x)) >= log(k) + log(d) || ab(x) >= k * d) {
		printf("%lld\n", ab(ab(x) - k * d));
		return 0;
	}
	ll n, m, c;
	if (x >= 0) {
		n = x % d;
		m = n - d;
		c = x / d;
	} else {
		n = (-x) % d;
		m = n - d;
		c = (-x) / d;
	}
	// printf("%lld %lld %lld\n", n, m, c);
	if (c <= k - 1) {
		if ((k - c) % 2 == 0) {
			printf("%lld\n", ab(n));
		} else {
			printf("%lld\n", ab(m));
		}
	} else {
		printf("%lld\n", ab(n));
	}
	return 0;
}