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)

// =====================================================
//   File    :   main.cpp
//   Author  :   eezyl
//   Synopsis:   codeforces/atcoder template
//   Address :
//   Date    :   2020/08/15
// =====================================================

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

#define rep(i, a, b) for (int i = (a); i < int(b); ++i)
typedef long long LL;

const int INF = (1<<31)-1;
const LL LL_INF = (1ll << 63)-1;
const int MOD = 1e9+7;

LL gcd (LL a, LL b) {
    if (b == 0)
        return a;
    return gcd(b, a%b);
}

void solve() {
    LL X, K, D;
    cin >> X >> K >> D;
    LL Y = abs(X);
    if (Y / D > K) {
        cout << Y-D*K << endl;
    } else {
        if (Y/D %2 == K%2) {
            cout << Y%D << endl;
        } else {
            cout << abs(Y%D-D) << endl;
        }
    }
}


int main()
{
    ios::sync_with_stdio(0);
    cin.tie(0);
//#define LOCAL
#ifdef LOCAL
freopen("data.in", "r", stdin);
while (!feof(stdin)) {
#endif // LOCAL

    int t = 1;
    //cin >> t;
    while (t--) {
        solve();
    }

#ifdef LOCAL
}
printf("Time used = %.2lf\n", (double)clock() / CLOCKS_PER_SEC);
#endif // LOCAL
    return 0;
}