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.

Examples

Input

6 2 4


Output

2


Input

7 4 3


Output

1


Input

10 1 2


Output

8


Input

1000000000000000 1000000000000000 1000000000000000


Output

1000000000000000
import java.io.*;
import java.util.Arrays;
import java.util.StringTokenizer;

public class Main {
    private static final BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    private static final PrintWriter pw = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
    private static StringTokenizer st;

    private static long readLong() throws IOException {
        while (st == null || !st.hasMoreTokens()) st = new StringTokenizer(br.readLine());
        return Long.parseLong(st.nextToken());
    }

    public static void main(String[] args) throws IOException {
        pw.println(solve());
        pw.close();
    }

    private static long solve() throws IOException {
        long OrigX = readLong();
        long K = readLong();
        long D = readLong();

        if (OrigX == 0) return D * (K % 2);

        return solve(Math.abs(OrigX), K, D);
    }

    private static long solve(long X, long K, long D) {
        if (X / D >= K) return (X - K * D);
            // Now X < D * K

        else {
            K -= X/D;
            X -= X / D * D;

            if ( K % 2 == 0) {
                return X;
            } else {
                return D - X;
            }
        }
    }
}
