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)

#pragma GCC optimize("O3")
#include <bits/stdc++.h>

using namespace std;

//defines
typedef long long ll;

using namespace std;

double pi=1.57079632679489661923;
ll modu=1e9+7;

ll t=1,n=1,q;

int main(){
    ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);

    ll x,k,d;
    cin>>x>>k>>d; x=abs(x);

    ll ans=x;

    ll low=0,hi=k,mid;
    ll eve, cur;

    while(low<=hi){
        mid = (low+hi)/2;

        cur = x - mid*d;
        eve = (k-mid)%2;

        ans = min( ans , min( abs(cur - eve*d), abs(cur + eve*d)  ) );

        if( 0<= cur ){ low = mid+1; }
        else{ hi = mid-1; }
    }

    mid++;

    cur = x - mid*d;
    eve = (k-mid)%2;

    ans = min( ans , min( abs(cur - eve*d), abs(cur + eve*d)  ) );

    cout<<ans<<endl;


}
/*
10
5 4
8 1 7 2
1 3
2 4
3 3
*/