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: Time Limit Exceeded)

#include <bits/stdc++.h>
#include <stdio.h>
#include <math.h>
#include <cmath>
using namespace std;
#define vsort(vec) sort(vec.begin(), vec.end())
#define vreverse(vec) reverse(vec.begin(), vec.end())
using ll =long long;
const ll INF =1001001001;

//最大公約数
ll gcd(ll a, ll b){
   if (a%b==0){
       return(b);
   }else{
       return(gcd(b,a%b));
   }
}
//最小公倍数
ll lcm(ll a, ll b){
   return a*b/gcd(a,b);
}



//階乗　20くらいが限界
ll factorial(ll n){
  ll a=1;ll b=n;
  for(int i=0;i<n;i++){
    a*=b;
    b--;
  }
  return a;
}

//組み合わせ　nCr
ll comb(ll a,ll b){
  ll c;
  c=factorial(a)/(factorial(a-b)*factorial(b));
  return c; 
}




/*10億7。　　　100000007
2*2行列 H*W
vector<vector<int>> vec(H,vector<int>(W));
for(int i=0;i<H;i++){
for(int j=0;j<W;j++){
cin>>vec[i][j];
  }
}

文字列→数列
string S;int A=atoi(S.c_str());
数列→文字列
int A;string S=to_string(A);
数字判定　isdigit(a)
pair宣言　pair<int,int> p(5,7);
ソート　sort(vec.begin(),vec.end());
キュー　queue<int> q;
   q.push(7);　//７が入る
   cout << q.front() << endl;  // 先頭の値を出力
   q.pop();  // 先頭の値を削除
*/



int main(void){
cout<<fixed<<setprecision(14);
//cin.tie(0); ios::sync_with_stdio(false);

ll X,K,D;
  cin>>X>>K>>D;
  ll point=X;
  for(int i=0;i<K;i++){
    if(abs(point-D)<abs(point+D)){
      point-=D;
    }else{
      point+=D;
    }
  }
    cout<<abs(point)<<endl;

  
  
    
}