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;

const int MOD = 1000000007;
const int INF = 1e9;

int main(){
  long long x;
  long long k, d;
  cin >> x >> k >> d;

  long long now = abs(x);
  long long ans = 0;

  long long cnt = 0;
  cnt = now / d;
  if(k >= cnt){
    now = now - d * cnt;
    cnt = k - cnt;
    if(cnt%2 == 0){
      ans = now;
    }else{
      ans = d - now;
    }
  }else{
    ans = now - k * d;
  }
  ans = abs(ans);
  // if(now >= 0){
  //   cnt = now / d;
  //   if(k >= cnt){
  //     now = now - d * cnt;
  //     cnt = k - cnt;
  //     if((k-cnt)%2 == 0){
  //       ans = now;
  //     }else{
  //       ans = d - now;
  //     }
  //   }else{
  //     ans = now - k * d;
  //   }
  // }else if(now < 0){
  //   now = -1 * now;
  //   cnt = now / d;
  //   if(k >= cnt){
  //     now = now - d * cnt;
  //     cnt = k - cnt;
  //     if((k-cnt)%2 == 0){
  //       ans = now;
  //     }else{
  //       ans = d - now;
  //     }
  //   }else{
  //     ans = now - k * d;
  //   }
  // }
  // ans = abs(ans);
  cout << ans << endl;

  return 0;
}