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)

/*
    ->->->->-> Vora Mahammasasim <-<-<-<-<-
*/
#include<bits/stdc++.h>
#include<stdio.h>
#define Quick std :: ios_base :: sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define int long long
#define pb emplace_back
#define Mod 1000000007
#define all(a) a.begin(),a.end()
#define ps(a) partial_sum(all(a),a.begin())
using namespace std;





signed main(){

    Quick;
    int x,k,d;
    cin>>x>>k>>d;
    x = abs(x);
    int p = x%d;
    int q = x/d;
    if(q == k) cout<<q;
    else if(q < k){
        k -= q;
        if(k&1) cout<<min(abs(p-d),abs(p+d));
        else cout<<p;
    }
    else cout<<x - (k*d);
}