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;
 
#define int             long long
#define ld              long double
#define fi              first
#define se              second
#define pb              push_back
#define mp              make_pair
#define fo(i,n)         for(i=0;i<n;i++)
#define fo1(i,n)        for(i=1;i<=n;i++)
#define foab(i,a,b)     for(i=a;i<=b;i++)
#define of(i,n)         for(i=n-1;i>=0;i--)
#define of1(i,n)        for(i=n;i>=1;i--)
#define ofab(i,a,b)     for(i=a;i>=b;i--)
#define sq(a)           (a)*(a)
#define all(x)          x.begin(),x.end()
#define rall(x)         x.rbegin(),x.rend()
#define sortall(x)      sort(all(x))
#define sortrall(x)     sort(rall(x))
#define clr(x)          memset(x,0,sizeof(x))
#define tr(it, a)       for(auto it=a.begin();it!=a.end();it++)
#define pii             pair<int,int>
#define mk(arr,n,type)  type *arr=new type[n];
#define vi              vector<int>
#define si              set<int>
#define usi             unordered_set<int>
#define mii             map<int,int>
#define umii            unordered_map<int,int>
#define pqmx            priority_queue<int>
#define pqmn            priority_queue<int,vi,greater<int>>
#define setbits(x)      __builtin_popcountll(x)
#define trailzrobits(x) __builtin_ctzll(x)
#define leadzrobits(x)  __builtin_clzll(x)
#define paritybits(x)   __builtin_parityll(x)
#define gcd             __gcd
#define lcm(x, y)       (x * y)/gcd(x,y)
#define endl            '\n'
#define sz(s)           (int)s.size()
#define sp(x,y)         fixed<<setprecision(y)<<x
const int mod = 1000000007;
const int inf = 1000000000000000000;
const ld PI = 3.1415926535897932384626;
const ld eps = 1e-12;
void __print(int32_t x) {cout<<x;}
void __print(int x) {cout<<x;}
void __print(double x) {cout<<x;}
void __print(ld x) {cout<<x;}
void __print(float x) {cout<<x;}
void __print(bool x) {cout<<(x?"true":"false");}
void __print(char x) {cout <<'\''<<x<<'\'';}
void __print(const char *x) {cout <<'\"' <<x<<'\"';}
void __print(const string &x) {cout<<'\"'<<x<<'\"';}
template<typename T,typename V>     void __print(const pair<T,V> &x) {cout<<'{';__print(x.first);cout<<',';__print(x.second);cout<<'}';}
template<typename T>                void __print(const T &x) {int f=0;cout<<'{';for(auto &i:x)cout<<(f++?",":""),__print(i);cout<<"}";}void _print(){cout<<"]\n";}
template <typename T,typename... V> void _print(T t, V... v) {__print(t);if(sizeof...(v))cout<<", ";_print(v...);}
#ifndef ONLINE_JUDGE
#define dbg(x...) cout<<"["<<#x<<"]=[";_print(x)
#else 
#define dbg(x...)
#endif
int power(int base,int exp) {int res=1;base%=mod;while(exp>0){if(exp&1)res=(res*base)%mod;base=(base*base)%mod;exp=exp>>1;}return res;}
int inv(int x) {return power(x,mod-2)%mod;}
// ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++




const int N=1e6+6,M=N;
int i,j;

void solve() {
    
    int x,k,d; cin>>x>>k>>d;
    x=abs(x);
    if(x>=k*d){
        cout<<(x-k*d);
        return;
    }
    int cnt=x/d;
    k-=cnt;
    int ans=x-cnt*d;
    
    if(k&1) ans=min(abs(ans-d),ans+d);
    cout<<ans;
}




int32_t main() {
    ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
    
    int t=1;
    // cin>>t;
    for(int i=1;i<=t;i++){
        // cout<<"case #"<<i<<": ";
        solve();
    }
    return 0;
}