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.util.*;

public class Main {
    public static void main(String[] args) throws Exception {
        // Your code here!
        Scanner sc = new Scanner(System.in);
        long now = sc.nextLong();
        long k = sc.nextLong();
        long d = sc.nextLong();
        
        long ans = 0;
        
        // 10 5 8
        // 2回行くことになる。 now = -6
        // 残り3回はプラスマイナスを往復 2,-6,2 ans = 2;
        long needcount = (Math.abs(now) + d+1) / d;
        // System.out.println(needcount);
        
        if (needcount > k) {
            if (now < 0) {
                ans = now + k * d;
                // System.out.println(ans);
            } else {
                ans = now - k * d;
                // System.out.println(ans);
            }
        } else {
            long leftcount = k - needcount;
            if (now < 0) {
                now = now + needcount * d;
            } else {
                now = now - needcount * d;
            }
            // System.out.println(now);
            
            if (leftcount % 2 == 1) {
                if (now < 0) {
                    now += d;
                } else {
                    now -= d;
                }
            }
            
            ans = now;
            // System.out.println(now);
        }
        // if (Math.abs(now) > Math.abs(k * d))
        System.out.println(Math.abs(ans));
    }
}
