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: Accepted)

#include <bits/stdc++.h>
using namespace std;

// 小数点以下桁数
// fixed << setprecision(i)

int64_t gcd(int64_t a,int64_t b){return (a%b==0?b:gcd(b, a%b));}
int64_t lcm(int a,int b){return (int64_t)a*b/gcd(a, b);}
int factorial(int a){int b=1;while(a){b*=a--;}return b;} //13以下対応
bool is_prime(int a){if(a<=1)return false;for(int i=2;(int64_t)i*i<=a;i++){if(a%i==0)return false;}return true;}
int get_adp(double x, int n){return (int)round((x-(int)x)*pow(10,n))%10;} // 小数点以下の指定桁の値を取得
int64_t sigma(int64_t s, int64_t n){return n*(2*s+n-1)/2;}

//順列の全列挙
// vector<int> v(N);
// iota(v.begin(), v.end(), 1);       // v に 1, 2, ... N を設定
// do {
//	for(auto x : v) cout << x << " "; cout << "\n";     // v の要素を表示
// } while( next_permutation(v.begin(), v.end()) ); 

//bit全探索
// for (int tmp = 0; tmp < (1 << ビット数); tmp++) {
//	bitset<ビット数> s(tmp);
//	// (ビット列sに対する処理)
// }

int main()
{
  int64_t x, k, d, ans;
  cin >> x >> k >> d;
  if (abs(x) / d > k)
  {
    if (x > 0)
      ans = x - k * d;
    else
      ans = abs(x + k * d);
  }
  else
  {
    if ((k - abs(x) / d) % 2 == 0)
      ans = abs(x) % d;
    else
      ans = abs(abs(x) % d - d);
    // ans = min(abs(x) % d, abs(abs(x) % d - d));
  }
  cout << ans << endl;
  
}