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>
using namespace std ;
#define rep(i,n) for( int i = 0 ; i < n ; i++ )
typedef long long ll ;

int main(){
    ll x , k , d ;
    cin >> x >> k >> d ;
    map<ll,ll> where ;
    where[x] = 1 ;
    ll now = x ;
    ll count = 1 ;
    ll ans = 1e15 + 1 ; 
    while( where[now]<=1 && k >= count ){
        if(abs( now - d ) > abs(now + d ) ) now = now + d ;
        else if (abs(now - d ) == abs(now + d ) ){
            if( ( k - count) %2 == 0 )cout << abs(d) <<endl ;
            else cout << abs(now) << endl ;
            return 0 ;
        }
        else now = now - d ;
        //cout << now << endl ;
        where[now]++ ;
        ans = min(abs(now),ans) ;
        count++ ;
    }
    cout << ans <<endl;
}