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>
#define repp(i,l,r)for(long long i=(l);i<(r);i++)
#define rep(i,n) for (long long i = 0; i < (n); ++i)
#define per(i,n) for (long long i = (n); i >= 0; --i)
#define all(v) v.begin(), v.end()
const int INF = 1 << 30;//int max
const long long int MOD = 1000000007;
using namespace std;
using ll = long long;
using P = pair<int,int>;
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; }
//ミョ(-ω- ?)
int main() {
    ll x,k,d;
    cin >> x >> k >> d;
    if(abs(x) >= k*d){
    //k回移動しても原点を通らないとき
        if(x < 0){
            x += k*d;
        }else x -= k*d;
        cout << abs(x) << "\n";

    }else{
        ll t = min(abs(x)/d,k);//予防
        if(x < 0)x += t*d;
        else x -= t*d;
        k -= t;
        //できるだけ原点に近づく

        if(k % 2 == 1){//反復横跳びすると跳んだ先で止まる
            if(x < 0)x += d;
            else x -= d;
            cout << abs(x) << "\n";
        }else{//もとに戻る
            cout << abs(x) << "\n";
        }
    }
    return 0;
}