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;

#define IOS                           \
    std::ios::sync_with_stdio(false); \
    cin.tie(NULL);                    \
    cout.tie(NULL);
 
#define rep(i, a, n) for(int i=a; i<n; i++)
#define ll long long
#define all(x) begin(x), end(x)
#define toInt(x) (x - '0')
#define pll pair<ll, ll>
#define pli pair<ll, int>
#define int long long
#define INF 1e+9
#define MAX_V 100008
const int MOD = 1e9 + 7;
const int inf = 1 << 30;



signed main(){
  IOS
  long long X, K, D;
  cin >> X >> K >> D;

  long long cnt = min(K, abs(X) / abs(D));
  long long curr = abs(X) - abs(cnt) * abs(D);
  long long amari = K - cnt;
  // cout << amari << endl;
  if(amari % 2 == 0){
    cout << curr << endl;
  }else{
    cout << abs(curr - D) << endl;
  }
  // cout << curr << endl;
  // if(K % cnt != 0){
  //   cout << curr << endl;
  // }else{
  //   cout << "k" << endl;
  // }

}