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;
#define ll  long long
#define fio ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL)
#define endl '\n'
#define mod 1000000007
int main() {
    fio;
    ll x, k, d;
    ll ans = 0;
    cin >> x >> k >> d;
    if (x <= 0)
    {
        ll n = (-1 * x) / d;
        ll temp = k - n;

        if (temp <= 0)
        {
            cout << abs(x + k * d) << endl;
            return 0;
        }
        x = x + n * d;
        k = k - n;
    }
    if (x < d)
    {
        x = x + d;
        k--;
    }
    ll steps = x / d;
    ll rem = k - steps;
    if (rem > 0 and rem % 2 == 0) {
        ans = abs(x % d);
    }
    else if (rem > 0 and rem % 2 != 0) {
        ans = abs((x % d) - d);
    }
    else if (rem <= 0) {
        ans = abs(x - k * d);
    }
    cout << ans;
    return 0;
}