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: Accepted)

#include <bits/stdc++.h>
#define int long long
#define gcd            __gcd
#define setbits(x)     __builtin_popcountll(x)
#define zrobits(x)     __builtin_ctzll(x)
#define mod            1000000007
#define mod2           998244353
#define maxe           *max_element
#define mine           *min_element
#define inf            1e18
#define deci(x,y)      fixed<<setprecision(y)<<x
#define w(t)           int t; cin>>t; while(t--)
#define nitin          ios_base::sync_with_stdio(false); cin.tie(NULL)
#define PI             3.141592653589793238
using namespace std;

int power(int x, int y, int p)
{
    int res = 1;
    x = x % p;
    while (y > 0)
    {
        if (y & 1)
            res = (res * x) % p;
        y = y >> 1;
        x = (x * x) % p;
    }
    return res;
}
int modi(int a, int m)
{
    return power(a, m - 2, m);
}
int32_t main()
{
    nitin;
    int cnt=0;
    int x,k,d;
    cin>>x>>k>>d;
    x=abs(x);
    int a=(x/d);
    int ans=abs((x-k*d));
    if( a<=k && (k-a)%2==0)
    {
        ans=min(ans,abs(x-a*d));
    }
    --a;
    if( a<=k && (k-a)%2==0)
    {
        ans=min(ans,abs(x-a*d));
    }
    ++a;
    if(a+1<=k && (k-(a+1))%2==0)
    {
        ans=min(ans,abs(x-(a+1)*d));
    }
    ++a;
    if(a+1<=k && (k-(a+1))%2==0)
    {
        ans=min(ans,abs(x-(a+1)*d));
    }
    cout<<ans<<endl;
    return 0;
}