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;

typedef long long ll;
typedef long double ld;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef pair<ll, ll> vll;
typedef pair<int, int> vii;

#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define ub upper_bound
#define lb lower_bound
#define asd cout << "ok" << endl;

#define all(x) x.begin(), x.end()
#define rall(x) x.rbegin(), x.rend()
#define trav(it, m) for(auto it = m.begin(); it!= m.end(); it++)

const char nl = '\n';
const int MX = 2e5 +5;
const int MOD = 1e9 +7;
const int inf = 1e9;
const ll INF = 1e18;
const ld PI = acos((ld)-1);

int fenwick[MX] = {};




int main(){
   ios_base::sync_with_stdio(0); cin.tie(0);
   ll x, k, d;
   cin >> x >> k >> d;
   ll tmp = x/d;
   k -= tmp;
   if(k%2) cout << abs(d-x%d) << nl;
   else cout << x%d << nl;

}