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 Java (OpenJDK 11.0.6) (judge verdict: Accepted)

import java.io.InputStream;
import java.io.PrintWriter;
import java.util.NoSuchElementException;
public class Main {
	public static void main(String[] args)throws Exception{
		PrintWriter out = new PrintWriter(System.out);
		long X=nextLong();
		long K=nextLong();
		long D=nextLong();
		X=Math.abs(X);
		long key=X/D;
		if(Math.abs(X-(D*key))>Math.abs(X-(D*(key+1))))
			key++;
		if(K>=key){
			K-=key;
			X=Math.abs(X-(key*D));
			if(K%2==1)
				System.out.println(Math.abs(X-D));
			else
				System.out.println(X);
		}
		else
			System.out.println(Math.abs(X-(D*K)));
	}
	public static long lmod(long i, long j) {
	     return (i%j)<0?(i%j)+0+(j<0?-j:j):(i%j+0);
	 }
	static boolean isOK(int index,int key,int[]a){
		if (a[index]>=key) return true;
	    else return false;
	}
	static int binary_search(int key,int[] a) {
	    int ng = -1; //「index = 0」が条件を満たすこともあるので、初期値は -1
	    int ok = (int)a.length; // 「index = a.size()-1」が条件を満たさないこともあるので、初期値は a.size()
	    while (Math.abs(ok - ng) > 1) {
	        int mid = (ok + ng) / 2;
	        if (isOK(mid,key,a)) ok = mid;
	        else ng = mid;
	    }
	    return ok;
	}
	//FastScanner
	static InputStream in = System.in;
	static  byte[] buffer = new byte[1024];
	static int length = 0, p = 0;
	public static boolean hasNextByte () {
		if (p < length) return true;
		else {
			p = 0;
			try {length = in.read(buffer);}
			catch (Exception e) {e.printStackTrace();}
			if (length == 0) return false;
		}
		return true;
	}
	public static int readByte () {
		if (hasNextByte() == true) return buffer[p++];
		return -1;
	}
	public static boolean isPrintable (int n) {return 33<=n&&n<=126;}
	public static void skip () {
		while (hasNextByte() && !isPrintable(buffer[p])) p++;
	}
	public static boolean hasNext () {skip(); return hasNextByte();}
	public static String next () {
		if (!hasNext()) throw new NoSuchElementException();
		StringBuilder sb = new StringBuilder();
		int temp = readByte();
		while (isPrintable(temp)) {
			sb.appendCodePoint(temp);
			temp = readByte();
		}
		return sb.toString();
	}
	public static int nextInt () {return Math.toIntExact(nextLong());}
	public static int[] nextInts (int n) {
		int[] ar = new int[n];
		for (int i=0; i<n; i++) ar[i] = nextInt();
		return ar;
	}
	public static long nextLong () {
		if (!hasNext()) throw new NoSuchElementException();
		boolean minus = false;
		int temp = readByte();
		if (temp == '-') {
			minus = true;
			temp = readByte();
		}
		if (temp<'0' || '9'<temp) throw new NumberFormatException();
		long n = 0;
		while (isPrintable(temp)) {
			if ('0'<=temp && temp<='9') {
				n *= 10;
				n += temp - '0';
			}
			temp = readByte();
		}
		return minus? -n : n;
	}
}