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;

long long int modPow(long long int a, long long int n){//繰り返し自乗法(a^n mod p)。計算量 O(logN)
    long long int p = pow(10, 9) + 7;//任意のp(素数) modのやつで使える 
    if(n == 0){
        return 1;
    }
    if(n == 1){
        return a % p;
    }
    if (n % 2 == 1){ 
        return (a * modPow(a, n - 1)) % p;
    }      
    long long t = modPow(a, n / 2);
    return (t * t) % p;
}

int main(){
    long long int x, k , d, ans;
    cin >> x >> k >> d;
    ans = 0;

    bool flag; // true:+. false:-
    long long int count = 0; 
    /*
    if(x > 0){
        //flag = true;
        while(x >= 0 && k > 0){
            x -= d;
            k--;
        }
    }else{
        //flag = false;
        while(x <= 0 && k > 0){
            x += d;
            k--;
        }
    }*/
    
    if(x == 0){

    }else if(x > 0){  
        if(x % d == 0){
            count  = x / d;
        }else{
            count = x / d + 1;
        }

        if(k <= count){
            x = x - k*d;
            k = 0;
        }else{
            x = x - d*count;
            k = k - count;
        }
    }else{
        if(x % d == 0){
            count  = abs(x) / d;
        }else{
            count = abs(x) / d + 1;
        }

        if(k <= count){
            x = x + k*d;
            k = 0;
        }else{
            x = x + d*count;
            k = k - count;
        }
    }


    if(k == 0){
        cout << abs(x) << endl;
    }else if(k % 2 == 0){
        cout << abs(x) << endl;
    }else{
        ans = min(abs(x + d), abs(x - d));
        cout << ans << endl;
    }
}