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>
#include <string>
#include <cstring>
#include <algorithm>
#include <math.h>
#include <map>
#include <bitset>
#include <cmath>
#include <set>
#include <utility>
#include <sstream>

#define INF 1000000000000
#define MOD 1000000007 //10^9+7:合同式の法

using namespace std;
typedef long long ll;

//マクロ
#define REP(i,n) for(ll i=0;i<(ll)(n);i++)
#define REPD(i,n) for(ll i=n-1;i>=0;i--)
#define FOR(i,a,b) for(ll i=a;i<=(ll)(b);i++)
#define FORD(i,a,b) for(ll i=a;i>=(ll)(b);i--)
#define ALL(x) x.begin(),x.end()
#define chmax(x,y) (x = max(x,y))
#define chmin(x,y) (x = min(x,y))

int main(){
    ll X,K,D;
    cin >> X >> K >> D;
    abs(X);
    ll syou,amari;
    syou = X/D;
    amari = X % D;
    ll res;
    if(syou == 1 and amari == 0 and (K % 2) == 0)
        cout << X << endl;
    else if(syou == 1 and amari == 0 and (K % 2) == 1)
        cout << amari <<endl;
    else if(syou <= K - 2)
        cout << amari << endl;
    else if(syou == (K -1)){
        res = abs(amari - D);
        cout << res << endl;
    }
    else{
        cout << abs(X - D * K) << endl;
    }

}