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)

//
//  main.cpp
//  ervb
//
//  Created by Kanak Gautam on 21/04/20.
//  Copyright © 2020 Kanak Gautam. All rights reserved.
//
 
 
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <cmath>
#include <vector>
#include <set>
#include <map>
#include <unordered_set>
#include <unordered_map>
#include <queue>
#include <ctime>
#include <cassert>
#include <complex>
#include <string>
#include <cstring>
#define pb push_back
#define mk make_pair
#define endl "\n"
#define mod 1000000007
using namespace std;
typedef long long int lli;
typedef long double ld;
priority_queue <lli, vector<lli>, greater<lli> > ti;
vector <lli> p[500005],g[500005],f(500005,0),b(500005,0),a(500005,0),ans(500005,0);
map <pair<lli,lli>,lli> mp;
set<pair<lli, lli> > s;
set<lli> st;
map<lli,lli> np,v;
queue<lli> qy;
lli n,q;
lli gcd(lli a, lli b)
{
    if (b == 0)
        return a;
    return gcd(b, a % b);
      
}
lli bpow(lli a, lli b) {
    lli res = 1;
    while (b > 0) {
        if (b & 1)
            res = (res * a)%mod;
        a = (a * a)%mod;
        b >>= 1;
    }
    return res%mod;
}
lli fact(lli i)
{
    f[0]=1;
    for(lli k=1;k<=i;k++)
    {
        (f[k]=f[k-1]*k)%=mod;
    }
    return f[i];
}
lli isprime(lli n)
{
    if(n==1)
        return 0;
    for(lli i=2;i<=sqrt(n);i++)
        if(n%i==0)
            return 0;
    return 1;
}
bool cmp(lli i,lli j)
{
    return p[i].size()<p[j].size();
}
/*void sieve()
{
    for(lli i=2;i<=5000000;i++)
    {
        if(b[i]==0)
        {
        for(lli j=2;i*j<=5000000;j++)
        {
            b[i*j]=1;
        }
        }
    }
}*/
int main ()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);cout.tie(NULL);
    lli x,k,d;
    cin>>x>>k>>d;
    x=abs(x);
    lli m=x/d;
    lli f=m;
    lli o=m+1;
    lli g=x-d*f;
    lli h=x-d*o;
    if(k%2==f%2 && k>=f)
    {
        cout<<abs(g)<<endl;
    }
    if(k%2 == o%2 && k>=o)
    {
        cout<<abs(h)<<endl;
    }
    else
    {
        cout<<abs(x-k*d)<<endl;
    }
}