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>
using namespace std;
using ll=long long;
template<class T,class U> inline bool chmin(T&x,U y){if(x>y){x=y;return true;}return false;}
template<class T,class U> inline bool chmax(T&x,U y){if(x<y){x=y;return true;}return false;}
#define fr(i,n) for(int i=0;i<(n);++i)
#define Fr(i,n) for(int i=1;i<=(n);++i)
#define ifr(i,n) for(int i=(n)-1;i>=0;--i)
#define iFr(i,n) for(int i=(n);i>0;--i)

int main(){
    cin.tie(nullptr);
    ios::sync_with_stdio(false);
    ll x,k,d;
    cin>>x>>k>>d;
    x=abs(x);
    if(k%2) --k,x-=d;
    x=abs(x);
    k/=2;
    d*=2;
    ll y=x/d,ans{};
    if(y<k) ans=min(x-d*y,d*y+d-x);
    else ans=x-d*k;
    cout<<ans<<'\n';
}