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)


// Problem : 
// 			C - Walking Takahashi
// 			Editorial
// 		
// Contest : AtCoder - AtCoder Beginner Contest 175
// URL : https://atcoder.jp/contests/abc175/tasks/abc175_c
// Memory Limit : 1024 MB
// Time Limit : 2000 ms
// Powered by CP Editor (https://github.com/cpeditor/cpeditor)

//#pragma GCC optimize "trapv"//  WA to RE int!!
#include <bits/stdc++.h>
#define FOR(i,n) for(int i=0;i<n;i++)
#define FORR(i,n) for(int i=n;i>=0;i--)
#define pb push_back
#define vlli vector<int>
#define slli set<int>
#define mlli map<int,int>
#define plii pair< int,int >
#define int long long int 
#define test int tt1234; cin>>tt1234;while(tt1234--)
#define endl "\n"
#define SI(a,n) int a[n];FOR(i,n) cin>>a[i];
#define input(a,n) int n;cin>>n; SI(a,n)
#define all(zz) zz.begin(),zz.end()
#define rep(i, begin, end) for (__typeof(end) i = (begin) - ((begin) > (end)); i != (end) - ((begin) > (end)); i += 1 - 2 * ((begin) > (end)))
#define error(x) cerr << #x <<" is " << x << endl;
#define Error(a,b) cerr<<"( "<<#a<<" , "<<#b<<" ) = ( "<<(a)<<" , "<<(b)<<" )\n";
#define precision(x) cout<<fixed<<setprecision(x);
#define fast ios_base::sync_with_stdio(false);cin.tie(NULL);
#define present(container, element) (container.find(element)!=container.end())
#define fint fast;test
#define set(a,x) memset(a,x,sizeof(a))
#define L(x)	((x)<<1)
#define R(x)	(((x)<<1)+1)
#define ft first
#define se second
#define MOD 1000000007
using namespace std;
template<typename... T> void rd(T&... args) { ((cin >> args), ...); }
template<typename... T> void pp(T... args) { ((cout << args << " "), ...); cout<<"\n"; }

void solve(){
	int x,k,d;
	rd(x,k,d);
	int N=(x)/d;
	if((x%d)==0)
	N-=1;
	if(N<0)N=0;
	if(N>=k){
		pp(x-k*d);
	}
	else pp(x-N*d);
	return;	
}


int32_t main() {
    // fint{
    	fast;
    	solve();
    // }
    return 0;
}