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;
 
typedef long long ll;
typedef long double ld;
typedef pair<int, int> Pii;
typedef pair<int, ll> Pil;
typedef pair<ll, ll> Pll;
typedef pair<ll, int> Pli;

#define fi first
#define se second

const ll MOD = 1e9 + 7;
const ll MOD2 = 998244353;
const ll MOD3 = 1812447359;
const ll INF = 1ll << 62;
const double PI = 2 * asin(1);

void yes() {printf("yes\n");}
void no() {printf("no\n");}
void Yes() {printf("Yes\n");}
void No() {printf("No\n");}
void YES() {printf("YES\n");}
void NO() {printf("NO\n");}

ll X, K, D;

int main(){
  cin >> X >> K >> D;
  ll cnt = abs(X) / D;

  if (K < cnt){
    if (X < 0){
      cout << abs(X + D * K) << endl;
    }else{
      cout << abs(X - D * K) << endl;
    }
    return 0;
  }

  ll now;
  if (X > 0) now = X - cnt * D;
  else now = X + cnt * D;

  ll ans = 1e18;
  if ((K - cnt) % 2 == 0){
    ans = min(ans, abs(now));
  }else{
    ans = min(abs(now + D), abs(now - D));
  }
  cout << ans << endl;

  return 0;
}