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>
#define pb push_back
#define mp make_pair
#define ff first
#define ss second
#define ll long long
#define ll_MAX LONG_LONG_MAX
#define ll_MIN LONG_LONG_MIN
#define pi pair<int,int>
#define endl "\n"
#define MAXN 100005
#define mod 1000000007
using namespace std;

void solve()
{
    
}

int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
	cout.tie(NULL);
    solve();
    ll x,k,d;
    cin>>x>>k>>d;
    if(x < 0)
    {
        if(x + k*d <= 0)
        {
            cout<<abs(x + k*d)<<'\n';
        }
        else
        {
            ll rem = k - abs(x)/d;
            x = x + (abs(x)/d)*d;
            if( (rem & 1) == 0)
            {
                cout<<abs(x)<<'\n';
            }
            else
            {
                cout<<abs(x + d)<<'\n';
            }  
        }
    }
    else if(x >= 0)
    {
        if( x - k*d >= 0)
        {
            cout<<abs(x - k*d)<<'\n';
        }
        else
        {
            ll rem = k - x/d;
            x = x - (x/d)*d;
            if( (rem & 1) == 0)
            {
                cout<<x<<'\n';
            }
            else
            {
                cout<<abs(x - d)<<'\n';
            }   
        }
    }
}