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 MOD 1000000007
//#define N 1000005
typedef long long int ll;
//const int mx=1e5+7;
//int n,m;
//vector<int> a(mx),b(mx);
// vector<int> g[mx];
// int deg[mx];
//vector<int> adj[mx];
//vector<int> ans;
//---dfs--
// bool visited[mx];
// vector<ll> val(mx,0);

// void dfs(int s,int e,int c1,int c2){
// 	if(a[s]^c1!=b[s]) {ans.push_back(s), c1^=1;}
// 	for(auto u:adj[s]){
// 		if(u!=e){
// 			dfs(u,s,c2,c1);
// 		}
// 	}
// }
//---dfs---

//---primes---
// const int N = 1e7 + 10;

// bool prime[N];
 
// int spf[N];
 
// void calc(){
// 	for(int i = 2; i<1e4; i++){
// 		if(prime[i])
// 			continue;
// 		spf[i] = i;
// 		for(int j = i * i; j < N; j += i){
// 			prime[j] = 1;
// 			spf[j] = i;
// 		}
// 	}
// 	for(int i = 2; i < N; i++)
// 		if(!spf[i])
// 			spf[i] = i;
// }

// vector<pair<int,int>> getp(int x){
// 	int curr=x;
// 	vector<pair<int,int>> ans;
// 	while(curr>1){
// 		int a=spf[curr];
// 		int count=0;
// 		while(curr%a==0){ curr/=a; count++;}
// 		ans.push_back(make_pair(a,count));
// 	}
// 	return ans;
// }

// bool ispr(int n) {
// 	if(n==1) return false;
//     for (int i=2;i*i<=n;++i){
//         if(n%i==0){
//             return false;
//         }
//     }
//     return true;
// }
//---primes---

//---segment_tree---
// int upd(int t,int val,int l,int r,int y, int z){
// 	if(y<l||y>r) return tr[val];
// 	if(l==r) return tr[val]=z;
// 	int s=(l+r)/2;
// 	upd(t^1,2*val+1,l,s,y,z);
// 	upd(t^1,2*val+2,s+1,r,y,z);
// 	if(t) tr[val]=tr[2*val+1]|tr[2*val+2];
// 	else tr[val]=tr[2*val+1]^tr[2*val+2];
// 	return tr[val];
// }
//int query(){}
//---segment_tree---

int main(){
	ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);cout<<setprecision(25);
	int T=1;
	//cin>>T;
	while(T--){
		ll x,k,d;
		cin>>x>>k>>d;
		x=abs(x);
		ll ans=0;
		if(x==0) {if(k%2==1) ans=d;}
		else{
			if(x>d){
				if(k<x/d){
					ans=x-k*d;
				}
				else{
					ll p=x-d*(x/d);
					ll q=k-x/d;
					if(q%2==1) ans=abs(p-d);
					else ans=p;
				}
			}
			else{
				if(k%2==1) ans=x-d;
				else ans=x;
			}
		}
		cout<<ans<<"\n";
 	}
}