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>
using namespace std;
using str = string;
using ll = long long;
#define REP(i,n) for(int i=0; i<int(n); i++)
#define RREP(i,n) for(int i=int(n)-1; i>=0; i--)
#define FOR(i,f,t) for(int i=int(f); i<=int(t); i++)
#define RFOR(i,f,t) for(int i=int(f); i>=int(t); i--)
#define ALL(vec) (vec).begin(),(vec).end()
#define ASORT(vec) sort(ALL(vec))
#define DSORT(vec) sort(ALL(vec), greater<int>());
#define MAX(x) *max_element(ALL(x))
#define MIN(x) *min_element(ALL(x))
#define YES(ans) if(ans) cout<<"YES"<<endl; else cout<<"NO"<<endl;
#define Yes(ans) if(ans) cout<<"Yes"<<endl; else cout<<"No"<<endl;
#define yes(ans) if(ans) cout<<"yes"<<endl; else cout<<"no"<<endl;

int main() {
    ll x, k, d, ans;
    cin >> x >> k >> d;
    if( k%2 == 1) {
        if(x>0) x = x - d;
        else x = x + d;
        k = k - 1;
    }
    if( 2*x/d >= 2*k+1 ) {
        ans = x - d * k;
    } else {
        if( x%(2*d) > d) ans = abs( 2*d - x%(2*d) );
        else ans = abs( x%(2*d) );
    }
    cout << ans << endl;
}