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 rep(begin, end) for (i=begin;i<end;i++)
#define repj(begin, end) for (j=begin;j<end;j++)
#define init(arr, end, val) for (i=0;i<end;i++) arr[i] = val;
#define printint(i0, i1) printf("%d %d\n", i0, i1)

using namespace std;
typedef long long int ll;
const int inf = 1000000000;
const int mod = 1000000007;
const int nil = -1;
ll i, n, m;
ll x, k, d, ans;

int main() {
    scanf(" %lld %lld %lld", &x, &k, &d);
    if (x == 0) {
        if (k%2 == 0) ans = 0;
        else ans = d;
    } else if (x > 0) {
        if (x - k * d > 0) ans = abs(x - k * d);
        else {
            i = floor(x / d);
            if (k%2 == i%2) ans = abs(x-i*d);
            else ans = abs(x - (i+1)*d);
        }
    } else{
        if (x + k * d < 0) ans = abs(x + k * d);
        else {
            i = floor((-x) / d);
            if (k%2 == i%2) ans = abs(x+i*d);
            else ans = abs(x + (i+1)*d);
        }
    }
    printf("%lld\n", ans);
}