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<string>
#include<math.h>
#include<queue>
#include<vector>
#include<map>
#include<iomanip>
#include<tuple>
#include<bitset>

using namespace std;
#define rep(i,n) for(int i=0;i<(n);i++)
#define printfloat(x) cout << fixed << setprecision(7) << (x) << endl;
#define printbit(x) cout << bitset<15>(x) << endl;
using ll = long long int;



long double pi = 3.14159265359;

long long int x,k,d,ans;


int main(){
    cin >> x >> k >> d;
    if(x <0){
        x = -x;
    }

    if(x >= 0 ){
        if(x-k*d > 0){
            ans = x-k*d;
        }
        else{
            // cout << k-(x+d)/d << " "  << endl;
            if((k-(x)/d) % 2){
                // cout << x-(x+d)/d*d+d << " ooooo" << endl;
                ans = x-(x)/d*d-d;
            }
            else{
                // cout << x-(x+d)/d*d <<  "1111" << endl;
                ans = (x-(x)/d*d);
            }
        }
    }
    // else{
    //     if(x+k*d < 0){
    //         ans = x+k*d;
    //     }
    //     else{

    //         if((k-(x-d)/d) % 2){
    //             ans = abs(x-(x+d)/d*d-d);
    //         }
    //         else{
    //             ans = abs(x-(x+d)/d*d);
    //         }
    //     }
    // }
    cout << ans << endl;

return 0;
}