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>
#include<string.h>
#include<vector>
#include<set>
#include<map>
#include<stack>
#include<queue>
#define endl  "\n"
#define ll long long
#define int long long
#define fast ios_base::sync_with_stdio(false);cin.tie(NULL);cin.exceptions(cin.failbit);cout.tie(NULL);
using namespace std;
bool isPrime(int n)
{
    if (n <= 1)
        return false;
    if (n <= 3)
        return true;
    if (n % 2 == 0 || n % 3 == 0)
        return false;
    for (int i = 5; i * i <= n; i += 6)
    {
        if ((n % i == 0) || (n % (i + 2) == 0))
            return false;
    }
    return true;
}

int32_t main()
{
	int t=1;
	while(t--)
	{
		int x,k,d;
		cin >> x >> k >> d;
		x=abs(x);
		
		int y;
		
		if(k>=x || d>=x)
		{
			y=-1;
		}
		else
		{
			int p=x/d;
			if(k>p)
			y=-1;
			else
			y=x-k*d;
		}
		
		
		if(y>=0)
		{
			cout << y;
		}
		else
		{
			
			if(x%d==0)
			{
				int temp=x/d;
				int final = k-temp;
				
				if(final%2==1)
				{
					cout << d;
				}
				else
				cout << "0";
			}
			else if(x>d)
			{
				int m=x%d;
				int temp=(x-m)/d;
				int final=k-temp;
				
				if(final%2==1)
				cout << d-m;
				else
				cout << m;
			}
			else
			{
				cout << min(x,abs(x-d));
			}
		}
	}
}