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)

#include<iostream>
#include<map>
#include<set>
#include<string>
#include<algorithm>
#include<cmath>
#include<vector>
#include<queue>
#include<stack>
#include<limits>
#include<sstream> 
#include<unordered_map>      

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef int itn;

const int inf = numeric_limits<int>::max() >> 2;
const ll linf = numeric_limits<ll>::max() >> 2;
const ull ulinf = numeric_limits<ull>::max() >> 2;
const double pi = acos(-1);
const int dx[4]={1,0,-1,0};
const int dy[4]={0,1,0,-1};
const int dx8[8]={-1,0,1,-1,1,-1,0,1};
const int dy8[8]={-1,-1,-1,0,0,1,1,1};

#define p_queue(i) priority_queue<i> 
#define rp_queue(i) priority_queue<i, vector<i>, greater<i>> 
#define umap(i,j) unordered_map<i,j>
#define P(p) cout<<(p)<<endl
#define PS(p) cout<<(p)<< " "
#define PN cout << endl
#define rep(i,m,n) for(int i = (m); i < (int)(n); i++)
#define rrep(i,m,n) for(int i = (n-1); i >= (int)(m); i--)
#define inrep(n,a) for(int i = 0; i < (int)(n); i++)cin >> a[i];
#define mod(i) ((i)%(ll)(1e9 + 7))
#define divm(a,b) (mod(a * modpow((ll)b,(ll)(1e9+5))))
#define rsort(a,b,c) sort(a, b, greater<c>())
#define vsort(v) sort(v.begin(), v.end())
#define rvsort(v,c) sort(v.begin(), v.end(), greater<c>()) 
#define ft first
#define sd second
#define pb push_back
#define it insert
#define sz(x) ((int)(x).size())
#define lb(a,n,k) (lower_bound(a,a+n,k) - (a)) 
#define vlb(a,k)  (lower_bound(a.begin(),a.end(),k) - a.begin())
#define ub(a,n,k) (upper_bound(a,a+n,k) - (a)) 
#define vub(a,k)  (upper_bound(a.begin(),a.end(),k) - a.begin())
#define YES cout<<"YES"<< endl
#define NO cout<<"NO"<<endl
#define Yes cout<<"Yes"<<endl
#define No cout<<"No"<<endl  
#define yes cout<<"yes"<<endl
#define no cout<<"no"<<endl
#define ret return

/*
ll modpow(ll i,ll j){ ll tmp=1; while(j){ if(j%2)tmp=mod(tmp*i);i=mod(i*i);j/=2;}return tmp;}
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a, ll b) { return a / gcd(a, b) * b; }
*/

/*
vector<string> split(const string &str, char sep){
    vector<string> v;
    stringstream ss(str);
    string buffer;
    while( getline(ss, buffer, sep) ) {
        v.push_back(buffer);
    }
    return v;
}
*/

/*
ll ncr[100][100];
ll nCr(int n, int r){if(n==r) ret ncr[n][r] = 1; if(r==0) ret ncr[n][r] = 1; if(r==1) ret ncr[n][r] = n;if(ncr[n][r]) ret ncr[n][r]; ret ncr[n][r] = nCr(n-1,r) + nCr(n-1,r-1);}

ll npr[100][100]={};
ll nPr(int n,int r){if(npr[n][r])ret npr[n][r];if(r==0)ret npr[n][r] = 1;if(r==1)ret npr[n][r] = n;ret npr[n][r] = n * nPr(n-1,r-1);}

ll nHr(int n,int r){ret nCr(n+r-1,r);}
*/

///////////////////////////////////////////////////////////////////////////



int main(){

    ll x,k,d;
    cin >> x >> k >> d;

    if(x < 0)x = -x;

    if(x/d >= k){
	x -= d*k;
	P(x);
	ret 0;
    }else{
	k -= x/d;
	x %= d;

	if(k&1){
	    P(min(abs(x+d),abs(x-d)));
	}else{
	    P(x);
	}
    }
    return  0;
}