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 <bits/stdc++.h>
 
using namespace std;

using usize = ::std::size_t;
using u64 = ::std::int_least64_t;
//using u64 = long long;
static constexpr u64 Inf = ::std::numeric_limits<u64>::max() / 2;


int main(int argc, char *argv[])
{
    
 
    cin.tie(0);
    ios::sync_with_stdio(false);
    
    u64 X, K, D;
    cin >> X >> K >> D;
    
    if (abs(X)/D > K) {
        u64 opt;
        if (X >= 0) {
            opt = X - K * D;
        } else {
            opt = X + K * D;
        }
        cout << opt << endl;
    } else {
        if ((abs(X)/D - K)%2 != 0) {
            u64 opt = X >= 0 ? X - (abs(X)/D)*D : X + (abs(X)/D)*D;
            u64 min_value = abs(opt + D);
            for (int i = -1; i <= 1; i += 2) {
                min_value = min(min_value, abs(opt + i * D));
            }
            cout << min_value << endl;
        } else {
            u64 opt = X >= 0 ? X - (abs(X)/D)*D : X + (abs(X)/D)*D;
            u64 min_value = abs(opt);
            for (int i = -2; i <= 2; i += 2) {
                min_value = min(min_value, abs(opt + i * D));
            }
            cout << min_value << endl;
        }
    }
    
    return 0;
}