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 <cstdlib>
#include <iostream>

using namespace std;

int main() {
    long long x, k, d;
    cin >> x >> k >> d;

    if (x > 0) {
        if (abs(x) / abs(d) > k) {
            cout << abs(x - d * k) << endl;
        } else {
            long long zero1 = x - abs(x) / abs(d) * d;
            long long zero2 = x - (abs(x) / abs(d) + 1) * d;
            if (abs(zero1) > abs(zero2)) {
                long long tyuusin = zero2;
                if ((k - abs(x) / abs(d) + 1) % 2 == 0) {
                    cout << abs(tyuusin) << endl;
                } else {
                    if (tyuusin > 0) {
                        cout << abs(tyuusin - d) << endl;
                    } else {
                        cout << abs(tyuusin + d) << endl;
                    }
                }

            } else {
                long long tyuusin = zero1;
                if ((k - abs(x) / abs(d) ) % 2 == 0) {
                    cout << abs(tyuusin) << endl;
                } else {
                    if (tyuusin > 0) {
                        cout << abs(tyuusin - d) << endl;
                    } else {
                        cout << abs(tyuusin + d) << endl;
                    }
                }
            }
        }
    } else {
        if (abs(x) / abs(d) > k) {
            cout << abs(x + d * k) << endl;
        } else {
            long long zero1 = x + abs(x) / abs(d) * d;
            long long zero2 = x + (abs(x) / abs(d) + 1) * d;
            if (abs(zero1) > abs(zero2)) {
                long long tyuusin = zero2;
                if ((k - abs(x) / abs(d) + 1) % 2 == 0) {
                    cout << abs(tyuusin) << endl;
                } else {
                    if (tyuusin > 0) {
                        cout << abs(tyuusin - d) << endl;
                    } else {
                        cout << abs(tyuusin + d) << endl;
                    }
                }

            } else {
                long long tyuusin = zero1;
                if ((k - abs(x) / abs(d) ) % 2 == 0) {
                    cout << abs(tyuusin) << endl;
                } else {
                    if (tyuusin > 0) {
                        cout << abs(tyuusin - d) << endl;
                    } else {
                        cout << abs(tyuusin + d) << endl;
                    }
                }
            }
        }
    }
}