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>

typedef uint64_t u64;
typedef int64_t i64;
typedef uint32_t u32;
typedef int32_t i32;
typedef uint16_t u16;
typedef int16_t i16;
typedef uint8_t u8;
typedef int8_t i8;

using namespace std;

#define rep(idx, N) for (u64 idx = 0; idx < N; idx++)

int main() {
  i64 x, k, d;

  cin >> x >> k >> d;

  i64 n1, n2;

  if (abs(x) / d >= k) {
    if (x > 0) {
      i64 ans = x;
      rep(i, k) ans -= d;
      cout << abs(ans);
      // cout << x - k * d;
      return 0;
    } else {
      i64 ans = x;
      rep(i, k) ans += d;
      cout << abs(ans);
      // cout << x + k * d;
      return 0;
    }
  }

  if (x > 0) {
    n1 = x - x / d * d;
    n2 = n1 - d;

  } else {
    n1 = abs(x) / d * d + x;
    n2 = n1 + d;
  }

  if ((abs(n1) <= abs(n2) && ((k - abs(x) / d) & 1) == 0) ||
      (abs(n1) >= abs(n2) && ((k - abs(x) / d) & 1) == 0)) {
    cout << abs(n1);
  } else {
    cout << abs(n2);
  }
  return 0;
}