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>
const double PI = acos(-1);
#define rep(i, n) for (int i = 0; i < (int)(n); i++ )
using namespace std;
int findSumOfDigits(int n) {
  int sum = 0;
  while (n > 0) { // n が 0 になるまで
    sum += n % 10;
    n /= 10;
  }
  return sum;
}
int main(){
    long long x,k,d,num,y;
    cin >> x >> k >> d;
    if(x < 0) x = -1*x;
    y = x;
    if(x == 0) cout << x << endl;
    else if( x < d ){
        if( 2*x <= d ) cout << x << endl;
        else cout << d - x << endl;
    }
    else if( x == d){
        if( k % 2 == 0 ) cout << x << endl;
        else cout << 0 << endl;
    }
    else if( x > 0){
        if( x >= k*d ){
            cout << x - k*d << endl;
        }
        else{
            rep(i,k){
                y -= d;
                if(y < 0){
                    num = i+1;
                    break;
                } 
            }   
            int ans = x - num*d; 
            for(int i = 0; i < k-num/2; i++){
                if(i % 2 == 0) ans += d;
                else ans -= d;
            }
            if(ans >= 0) cout << ans << endl;
            else cout << -1*ans << endl;
        }   
    } 
    
}