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: Wrong Answer)

#include <iostream>
#include <stdio.h>
#include <string>
// #include <cstdlib>
#include <stdlib.h>
using namespace std;

long long X;
long long K;
long long D;


int main(){
    cin >> X >> K >> D;

    if(D > abs(X)){
        if(K % 2 == 0){
            cout << abs(X) << endl;
        }
        else{
            if(X >= 0){
                cout << abs(X - D) << endl;
            }
            else{
                cout << abs(X + D) << endl;
            }
        }
    }
    else{
        if(D * K < abs(X)){
            if(X >= 0){
                cout << X - (D * K) << endl;
            }
            else{
                cout << abs(X + (D * K)) << endl;
            }
        }
        else{
            if(X >= 0){
                while(X >= 0 && K > 0){
                    X -= D;
                    K--;
                }
                if(K % 2 == 0){
                    cout << abs(X) << endl;
                }
                else{
                    cout << X + D << endl;
                }
            }
            else{
                while (X <= 0 && K > 0){
                    X += D;
                    K--;
                }
                if (K % 2 == 0)
                {
                    cout << abs(X) << endl;
                }
                else
                {
                    cout << abs(X - D) << endl;
                }
            }
        }
    }


}