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>
//     #include<boost/multiprecision/cpp_int.hpp>
// using boost::multiprecision::cpp_int; // use cpp_ll as data type for large numbers 
    using namespace std;
    typedef long long ll;
    typedef long double ld;
    typedef unsigned long long ull;
    typedef vector<int> vi;
    typedef vector<long long> vll;
    typedef pair<ll,ll> pll;
     
    const ll mod = 1e9 + 7;
    const ld PI = 3.1415926535897932384626;
    const ld eps = 1e-9;                            
     
    #define boost ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL)
    #define sidha(i,a,b) for(ll i=(ll)(a);i<(ll)(b);++i)
    #define ulta(i,a,b) for(ll i=(ll)(a);i>=(ll)(b);--i)
    #define stl(container,itr) \
        for(auto itr = container.begin(); itr != container.end(); itr++)
    #define sab(v) v.begin(), v.end()
    #define rall(v) v.rbegin(), v.rend()            
    #define test ll t;cin>>t;while(t--)
    #define sz size()
    #define pb push_back
    #define mp make_pair
    #define fi first
    #define se second
    #define LOG2(X) (63ll - __builtin_clzll(X))
    #define endl '\n'
    #define modinv(x) power(x, mod-2)
    #define MAX 100005
     
    ll power(ll x, ll y) { 
        ll ans=1;
        x=x%mod;
        while(y>0) { 
            if (y&1) 
                ans=(ans*x)%mod; 
            y=y>>1;
            x=(x*x)%mod; 
        } 
        return ans; 
    }
    
    // vector<ll> arr[MAX];
    // ll visit[MAX];
    // void dfs(ll n){
    //     visit[n]=1;
    //     for(ll i=0;i<arr[n].size();++i){
    //         if(visit[arr[n][i]]==0) dfs(arr[n][i]);
    //     }
    // }
    
    // ll spf[MAX];
    // void sieve(){
    //     sidha(i,1,MAX)  spf[i]=i;
    //     for(ll i=4;i<MAX;i+=2)  spf[i]=2;
    //     for(ll i=3;i*i<=MAX;i++){
    //         if(spf[i]==i){
    //             for(ll j=i*i;j<MAX;j+=i){
    //                 if(spf[j]==j)   spf[j]=i;
    //             }
    //         }
    //     }
    // }

    int main()
    {
        boost;
        ll x,k,d;
        cin>>x>>k>>d;
        if(x>0){
            // if(x-k*d>=0)    cout<<x-k*d;
            // else{
                ll tmp=x/d;
                tmp=min(tmp,k);
                ll tmp1=min(tmp,k)*d;
                tmp1=x-tmp1;
                if((k-tmp)%2)   cout<<abs(tmp1-d);
                else    cout<<abs(tmp1);
            //}
        }
        else if(x<0){
            // if(x+k*d<=0)    cout<<abs(x+k*d);
            // else{
                ll tmp=(abs(x))/d;
                tmp=min(tmp,k);
                ll tmp1=min(tmp,k)*d;
                tmp1=x+tmp1;
                if((k-tmp)%2)   cout<<abs(tmp1+d);
                else    cout<<abs(tmp1);
            //}
        }
        else{
            if(k%2) cout<<d;
            else    cout<<"0";
        }
        return 0;
    }