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 ll         = long long;
constexpr ll MOD = 1000000007;
#define REP(i, n) for (ll i = 0; i < (n); i++)
#define REP2(i, x, n) for (ll i = x; i < (n); i++)
const long long INF = numeric_limits<long long>::max();
template <class T> inline bool chmin(T& a, T b) {
    if (a > b) {
        a = b;
        return true;
    }
    return false;
}
template <class T> inline bool chmax(T& a, T b) {
    if (a < b) {
        a = b;
        return true;
    }
    return false;
}
vector<ll> getNums(ll N) {
    vector<ll> num(N);
    REP(i, N) {
        cin >> num[i];
    }
    return num;
}
int main() {
    ll X, K, D;
    cin >> X >> K >> D;

    ll ans = INF;

    ll begin = max(min(K / 2, abs(X) / D) - 10000000, (ll)0);
    if (K % 2 == 0) {
        if (X >= 0) {
            for (ll i = begin; i <= K / 2; ++i) {
                ll tmp = X - D * 2 * i;
                chmin(ans, abs(tmp));
                if (tmp < 0) break;
            }
        } else {
            for (ll i = begin; i <= K / 2; ++i) {
                ll tmp = X + D * 2 * i;
                chmin(ans, abs(tmp));
                if (tmp > 0) break;
            }
        }
    } else {
        if (X >= 0) {
            for (ll i = begin; i <= (K) / 2; ++i) {
                ll tmp = X - D * (i * 2 + 1);
                chmin(ans, abs(tmp));
                if (tmp < 0) break;
            }
        } else {
            for (ll i = begin; i <= (K) / 2; ++i) {
                ll tmp = X + D * (i * 2 + 1);
                chmin(ans, abs(tmp));
                if (tmp > 0) break;
            }
        }
    }

    cout << ans << endl;

    return 0;
}