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 <algorithm>
#include <iostream>
#include <iomanip>
#include <cassert>
#include <cstring>
#include <string>
#include <vector>
#include <random>
#include <bitset>
#include <queue>
#include <cmath>
#include <unordered_map>
#include <set>
#include <map>
#define INCANT cin.tie(0), cout.tie(0), ios::sync_with_stdio(0), cout << fixed << setprecision(20);
#define rep(i,n) for (int i=0; i<n;++i)
#define ALL(a)  (a).begin(),(a).end()
#define PI 3.14159265358979
typedef long long ll;
using namespace std;
const ll MOD = 1e9+7LL;
const int INF = 2e9;
ll X, K, D;

int main() {
  INCANT;
  cin >> X >> K >> D;
  X=abs(X); // 絶対値なので関係なし
  if (X-K*D >= 0) {
    cout << X-K*D << endl;
    return 0;
  }
  K-=X/D;
  X%=D;
  if (K%2==1) {
    cout << abs(X-D)  << endl;
  } else cout << X  << endl;
  return 0;
}