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)

//#pragma GCC optimize ("Ofast")
//#pragma GCC optimize ("unroll-loops")
//#pragma GCC target("avx,avx2,fma")

#include <bits/stdc++.h>
#include<ext/pb_ds/tree_policy.hpp>
#include<ext/pb_ds/assoc_container.hpp>

#define ios ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0)
#define file(s) if (fopen(s".in", "r")) freopen(s".in", "r", stdin), freopen(s".out", "w", stdout)
#define all(a) a.begin() , a.end()
#define F first
#define S second

using namespace std;
using ll = long long;
using namespace __gnu_pbds;

template<typename T> using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;

const int N = 2e5 + 10 , inf = 2e9 + 7;
const ll INF = 1e18 ,   mod = 1e9+7;

void solve() {
    ll x , k , d;
    cin >> x >> k >> d;
    x = abs(x);
    if(k <= x/d) {
        cout << x-d*k;
    }
    else {
        ll movs = k - (x/d);
        if(movs % 2 == 0) {
            cout << x%d;
        }
        else {
            cout << abs(x%d - d);
        }
    }
}

/*
1 1 5 5
1 2 13
*/
main() {
    ios;
    int tt = 1;
   // cin >> tt;
    while(tt --) {
        solve();
    }
    return 0;
}