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;
#define pb push_back
#define gcd __gcd
#define ff first
#define ss second
#define pii pair<int, int>
#define pll pair<ll, ll>
#define vi vector<int>
#define vec(cr)vector<cr>
#define fr(j, n)for(int i=j; i<n; i++)
#define scan(cr)for(auto &it:cr)cin >> it
#define all(cr)cr.begin(), cr.end()
#define sz(cr)cr.size()
#define Test() int t; for(cin >> t; t>0; t--)
#define deb(x)cout << #x << ":" << x << '\n'
int lcm(int a, int b){ return a/gcd(a, b) * b; }

int main(){
	ll n, k, d;
	cin >> n >> k >> d;
	ll moves = min(n/d, k);
	n -= (moves*d);
	k -= moves;
	if(k%2==0)cout << n;
	else cout << abs(n-d);
	return 0;
}