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)

//g++ t.cpp -o t && t < p.txt
//d>p.txt&&t<p.txt&&t2<p.txt

#include <iostream> // プログラムの耳と口
#include <vector> // 超有能配列秘書
#include <algorithm> // みんな大好きソートと二分探索
#include <queue> // きゅーちゃん、だいすき
#include <string> // to_string
#include <functional> // 関数を変数に入れる子
#include <set> // 値の取得・挿入・削除を高速に
#include <map> // setの妹 これまた優秀
#include <random> // 乱択さん
#include <ctime>
#include <bitset>

#define DB cerr<<"D"<<endl
using namespace std; using ll=long long; using ld=long double; const int INF=1e9; const ll LINF=1e18; const double dINF = 1e18; const ld ldINF = 1e18; const double EPS = 1e-6;
template<typename T, typename U, typename O> void caut(T a, U b, O c){cout<<"("<<a<<","<<b<<","<<c<<") ";} template<typename T, typename U> void caut(T a, U b){cout<<"("<<a<<","<<b<<") ";} template<typename T> void caut(T a){cout<<"("<<a<<") ";};
using P=pair<ll,ll>;
const ll M = 1e9+7;
ll mod_pow(ll x, ll a) { ll an = 1; while(a > 0) { if (a&1) an = an * x % M; x = x * x % M; a >>= 1;} return an;}
void add(ll& x, ll y) {x+=y; x%=M;}; void mul(ll& x, ll y) {x*=y; x%=M;}; template<typename T, typename U> void chmax(T& x, U y) {if (x<y) x=y;}; template<typename T, typename U> void chmin(T& x, U y) {if (x>y) x=y;}
bool vaild(int x, int y, int hh, int ww){return 0<=x&&x<hh&&0<=y&&y<ww;}
int keta(ll a) {int res=0; while(a>0) res+=a%10, a/=10; return res;}
const int up[]={1,-1,0,0}, lf[]={0,0,1,-1};

int main() {
  ll x,k,d; cin>>x>>k>>d;
  x=abs(x);
  if (x >= k*d) {
    cout<<x - k*d<<endl;
    return 0;
  }
  k -= x/d;
  x %= d;
  if (k%2 == 0) {
    cout<<x<<endl;
  } else {
    cout<<abs(x-d)<<endl;
  }
}
// ・配列の大きさok？　・priority_queueはgreater？　・debug消した？
// ・落ち着いて。提出まで4分待ってね……WJ……1/10……2/10……

// Thank you for making problems and running the contest