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>
#pragma GCC optimize("Ofast")
using namespace std;
typedef long long ll;
typedef pair<ll, ll> PLL;
typedef vector<int> VI;
typedef vector<char> VC;
typedef vector<double> VD;
typedef vector<string> VS;
typedef vector<ll> VLL;
typedef vector<PLL> VP;
const static int INF = 1000000000;
const static int MOD = 1000000007;
const int dx[4]={1, 0, -1, 0};
const int dy[4]={0, 1, 0, -1};
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }
#define rep(i,n) for (ll i=0; i<(ll)(n); i++)
#define repd(i,n) for (ll i=n-1; i>=0; i--)
#define rept(i,m,n) for(ll i=m; i<n; i++)
#define stl_rep(itr, x) for (auto itr = x.begin(); itr != x.end(); ++itr)
#define all(x) (x).begin(), (x).end()
#define F first
#define S second
#define PF push_front
#define PB push_back
#define SORT(V) sort((V).begin(), (V).end())
#define RVERSE(V) reverse((V).begin(), (V).end())
#define paired make_pair
#define PRINT(V) for(auto v : (V)) cout << v << " "
//charを整数に
int ctoi(char c) { if (c >= '0' && c <= '9') { return c - '0'; } return 0; }
// 累積和 for (int i = 0; i < N; ++i) s[i+1] = s[i] + a[i]; 
void cum_sum(int N,vector<double> a, vector<double> &s){ for(int i=0; i<N; i++){ s[i+1]=s[i]+a[i];}}
//ユークリッドの控除法
ll gcd(ll a,ll b){
    if(b == 0) return a;
    return gcd(b,a%b);
}
//最小公倍数
ll lcm(ll a,ll b){
    ll g = gcd(a,b);
    return a / g * b; // Be careful not to overflow
}
//素数判定
bool is_prime(long long n) {
    if (n <= 1) return false;
    for (long long p = 2; p * p <= n; ++p) {
        if (n % p == 0) return false;
    }
    return true;
}
int getdigit(ll num){
    unsigned digit=0;
    while(num!=0){
        num /= 10;
        digit++;
    }
    return digit;
}
//空白文字も入力 getline(cin, S);
//桁数指定 setprecision

int main()
{
    ios::sync_with_stdio(0); 
    cin.tie(0); 
    cout.tie(0);
    ll X, K, D;
    cin >> X >> K >> D;
    ll times=X/D;
    ll distance=X%D;
    if(times>=K){
        cout << distance+(times-K)*D << endl;
        return 0;
    }
    K-=times;
    ll ans1=0;
    ll ans2=0;
    if(K%2==0){
        ans1=distance;
    }
    else{
        ans1=abs(distance-D);
    }
    cout << ans1 << endl;
}