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)

#pragma GCC optimize("Ofast")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,fma")
#pragma GCC optimize("unroll-loops")
#include<bits/stdc++.h>
#define pii pair<int,int>
#define fi first
#define se second
#define int long long
#define pb push_back
#define mod 1000000007
#define pll pair<long long,long long>
using namespace std;
long long power(long long a,long long b){
    long long ans=1;
    while(b>0){
        if(b&1){ans=(ans*a)%mod;}
                a=(a*a)%mod;
                b>>=1;
        }
    return ans;
        
}
int find(int i,int a[]){
	if(a[i]==i)return i;
	return a[i]=find(a[i],a);
}
void uni(int i,int j,int a[],int sz[]){
	int x=find(i,a);
	int y=find(j,a);
	if(sz[x]>sz[y])a[y]=x;
	else a[x]=y;
	sz[x]+=sz[y];
	sz[y]=sz[x];
}

int32_t main(){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    int k,x,d;
    cin>>x>>k>>d;
    x=abs(x);
    int y=x/d;
    if(y>k){
    	cout<<x-k*d;
    	return 0;
    }
    if(y%2==k%2)cout<<x-y*d;
    else cout<<abs(x-(y+1)*d);
    return 0;
}