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: Accepted)

#include<bits/stdc++.h>
#define pb push_back
#define mp make_pair
#define fir first
#define sec second
#define SZ(a) (int)((a).size())
#define all(x) (x).begin(), (x).end()
using namespace std;
typedef long long ll;
typedef double db;
typedef pair <int, int> Pii;
typedef pair <ll, int> Pli;
typedef unsigned long long ull;
const double eps = 1e-8;
const int mod = (int)(1e9) + 7;
const int inf = 0x7fffffff;
const int N = (int)(2e5) + 7;

ll x, k, d;

int main() {

    // freopen("in.in", "r", stdin);
    scanf("%lld%lld%lld", &x, &k, &d);
    x = abs(x);
    if (x % d) {
        if (k < (x + d - 1) / d) printf("%lld\n", x - k * d);
        else {
            ll d1 = x % d, d2 = (d - x % d) % d;
            k -= (x + d - 1) / d;
            if (k & 1) printf("%lld\n", d1);
            else printf("%lld\n", d2);
        }
    } else {
        if (k <= x / d) printf("%lld\n", x - k * d);
        else {
            k -= x / d;
            if (k & 1) printf("%lld\n", d);
            else puts("0");
        }
    }

    return 0;
}