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 <bits/stdc++.h>

using namespace std;

const int N = 5e5+5;

const int mod = 1e9+7;

long long x , k , d;

int main(){
    cin >> x >> k >> d;
    long long ans = -1;
    if(x > 0){
        //first +ve number
        //10^15
        long long toP = x - min((x - 0) / d , k) * d;
        long long tmpK = (k - ((x-0)/d)) % 2;
        if(tmpK){
            toP = min(abs(toP + d) , abs(toP - d));
        }
        //first -ve number
        long long toN = x - min((x - 0 + d-1) / d , k) * d;
        long long tmpK2 = (k - ((x-0+d-1)/d)) % 2;
        if(tmpK2){
            toN = min(abs(toN - d) , abs(toN + d));
        }

        ans = min(abs(toN) , abs(toP));
    } else {
        //first -ve number
        long long toN = x + min((0 - x) / d , k) * d;
        long long tmpK = (k - ((0-x)/d)) % 2;
        if(tmpK){
            toN = min( abs(toN + d) , abs(toN - d));
        }
        //first +ve number
        long long toP = x + min((0 - x + d-1) / d , k) * d;
        long long tmpK2 = (k - ((0-x+d-1)/d)) % 2;
        if(tmpK2){
            toP = min(abs(toP + d) , abs(toP - d));
        }
        ans = min(abs(toN) , abs(toP));
    }
    cout << ans << "\n";
}