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 <cstdio>
#include <cmath>
#include <limits>
#include <fstream>
#include <iostream>
#include <algorithm>
#include <vector>
#include <utility>
#include <cstring>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <cstdint>
#define rep(i, s, g) for (i = s; i < g; i++)
using namespace std;
using ll = long long;

const ll mod = 1000000000 + 7; //10^9

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


    bool c = false;
    
    
    for (ll i = 0; i < K; i++)
    {
        if (X >= 0)
        {
            X -= D;
            if (X < 0)
            {
                if (D % 2 == 1)
    {
        c = true;
    }
    
    if (K > 5)
    {
        while (K > 5)
        {
            K /= 2;
        }
        if (K % 2 == 0 && c)
        {
            K++;
        }
        else if(K % 2 == 1 && !c){
            K++;
        }
        
    }
            }
            
            
            
        }else
        {
            X += D;
            if (X > 0)
            {
                if (D % 2 == 1)
    {
        c = true;
    }
    
    if (K > 5)
    {
        while (K > 5)
        {
            K /= 2;
        }
        if (K % 2 == 0 && c)
        {
            K++;
        }
        else if(K % 2 == 1 && !c){
            K++;
        }
        
    }
            }
            
        }
            // cout << X << endl;

    }
    cout << abs(X) << endl;
    
    
}