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 endl "\n"
#define fo for(int i=0;i<n;i++)
#define clear cin.clear();cin.ignore(numeric_limits<streamsize>::max(), '\n');
#define loopSolve int t; cin>>t; while(t--){solve();}
#define rep(i, n)    for(int i = 0; i < (n); ++i)
#define repA(i, a, n)  for(int i = a; i <= (n); ++i)
#define repD(i, a, n)  for(int i = a; i >= (n); --i)
#define trav(a, x) for(auto& a : x)
#define all(x) x.begin(), x.end()
#define sz(x) (int)(x).size()
#define fill(a)  memset(a, 0, sizeof (a))
#define fst first
#define snd second
#define mp make_pair
#define pb push_back
typedef long double ld;
typedef long long ll;
typedef pair<int, int> pii;
typedef vector<int> vi;

//INT_MAX         2,147,483,647                       <10^10
//U_INT_MAX       4,294,967,295                       <10^10
//LONG_LONG_MAX   9,223,372,036,854,775,807           <10^18
//U_LONG_LONG_MAX 18,446,744,073,709,551,615          <10^18
//ll power(int a, int b){
//    return (ll)(pow(a,b)+0.5);
//}

vector<int> gV(int n){
    vector<int> v(n);
    fo { cin>>v[i];  }
    return v;
}

int* gA(int n){
    int* arr = new int[n];
    fo cin>>arr[i];
    return arr;
}

bool isPossible(ll val,ll target,ll k, ll d){
    return (abs(target-val)%d==0)&&((abs(target-val)/d) ==k);
}


void solve(){

    ll val,k,d; cin>>val>>k>>d;

    val = abs(val);

    ll maxSteps = (val)/d;
    if(val%d==0) maxSteps--;

    if(maxSteps<k){
        ll curr = abs(val-(maxSteps*d));
        ll temp = curr;
        ll currSteps=0;
        while(abs(curr-d)<abs(curr) && currSteps+maxSteps<=k){
            curr-=d;
            currSteps++;
        }
        if((k-maxSteps-currSteps)%2)
            curr-=d;

        cout<<min(abs(curr),temp)<<endl;
    }
    else{
        cout<<abs(val-(k*d))<<endl;
    }
//    val = abs(val);
//
//    ll ans=val;
//
//    ll begin = val-k*d, end = val;
//
//    while(begin<=end){
//        ll mid = (begin+end)/2;
//
//        cout<<begin<<" "<<mid<<" "<<end<<endl;
//        if(isPossible(val,mid,k,d) || isPossible(val,-mid,k,d)){
//            end = mid-1;
//            ans=min(ans,abs(mid));
//        }else{
//            end = mid-1;
//        }
//    }


}


int main() {
	ios_base::sync_with_stdio(false);
    std::cin.tie(NULL);
    std::cout.tie(NULL);

    solve();
    //loopSolve;

	return 0;
}