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 int ll;

#define FOR(i, a, b) for (int i=a; i<(b); i++)
#define For(i, a) for (int i=0; i<(a); i++)
#define FORR(i,a,b) for (int i = (b)-1; i >= a; i--)
#define Forr(i,a) for (int i = (a)-1; i >= 0; i--)

#define deb(...) cerr << "[" << #__VA_ARGS__ << "] : [", DBG(__VA_ARGS__) 
void DBG() { cerr << "]\n";}
template<typename T, typename ...Args> void DBG(T first, Args... args){
cerr << first; if(sizeof...(args))cerr<<", "; DBG(args...); }

#define int ll
#define pb(x) push_back(x)
#define INF (int)1e17
#define PI 3.14159265358979323846
 
int del_x[4] = {-1, 1, 0, 0};
int del_y[4] = {0, 0, 1, -1};
int XX[8] = {-1, -1, -1, 0, 1, 1, 1, 0};
int YY[8] = {-1, 0, 1, 1, 1, 0, -1, -1};


void solve(){
	int x,k,d; cin >> x >> k >> d;
	int n = x/d;
	if(n>k)n=k;
	int rem = k-n;
	int final = x-n*d;
	if(n+1<=k && abs(x-n*d-d)<x-n*d){
		final = abs(x-n*d-d);
		rem--;
	}
	if(rem&1){
		cout << min(abs(final-d),abs(final+d))<<"\n";
	}
	else{
		cout << final <<"\n";
	}
	

}

int32_t main(){ 
	ios_base:: sync_with_stdio(false); 
	cin.tie(0); 
	int t=1;
	//cin >> t;
	for(int tt=1;tt<=t;tt++){
		//cout << "Case #"<<tt<<": ";
		solve();
	}
}