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++ (Clang 10.0.0) (judge verdict: Wrong Answer)

#include <algorithm>
#include <cmath>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <utility>
#include <vector>
#define rep(i, n) for (int i = 0, i##_len = (n); i < i##_len; ++i)
#define reps(i, n) for (int i = 1, i##_len = (n); i <= i##_len; ++i)
#define rrep(i, n) for (int i = ((int)(n)-1); i >= 0; --i)
#define rreps(i, n) for (int i = ((int)(n)); i > 0; --i)
#define INF 1000000000000
#define MOD 1000000007
#define CHOOSE(a) CHOOSE2 a
#define CHOOSE2(a0, a1, a2, a3, a4, x, ...) x
#define debug_1(x1) cout << #x1 << ": " << x1 << endl
#define debug_2(x1, x2) \
    cout << #x1 << ": " << x1 << ", " #x2 << ": " << x2 << endl
#define debug_3(x1, x2, x3)                                                 \
    cout << #x1 << ": " << x1 << ", " #x2 << ": " << x2 << ", " #x3 << ": " \
         << x3 << endl
#define debug_4(x1, x2, x3, x4)                                             \
    cout << #x1 << ": " << x1 << ", " #x2 << ": " << x2 << ", " #x3 << ": " \
         << x3 << ", " #x4 << ": " << x4 << endl
#define debug_5(x1, x2, x3, x4, x5)                                         \
    cout << #x1 << ": " << x1 << ", " #x2 << ": " << x2 << ", " #x3 << ": " \
         << x3 << ", " #x4 << ": " << x4 << ", " #x5 << ": " << x5 << endl
#ifdef _DEBUG
#define debug(...)                                                        \
    CHOOSE((__VA_ARGS__, debug_5, debug_4, debug_3, debug_2, debug_1, ~)) \
    (__VA_ARGS__)
#else
#define debug(...)
#endif

using namespace std;
using ll = long long;

int main() {
    ll x, k, d;
    cin >> x >> k >> d;
    // 最小値にいけない時
    if (x / k >= d) {
        cout << x - k * d << endl;
        return 0;
    }
    // 最小値にいける時
    ll mi = x % d;
    if (abs(k - int(x / d)) % 2 == 1)
        cout << abs(mi - d) << endl;
    else
        cout << mi << endl;

    return 0;
}