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 region header
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define rep1(i, n) for (int i = 1; i <= (int)(n); i++)
#define rev(i, n) for(int i = (int)(n - 1); i >= 0; i--)
#define rev1(i, n) for(int i = (int)(n); i > 0; i--)
#define pb push_back
#define all(v) (v).begin(), (v).end()
#define resort(v) sort((v).rbegin(), (v).rend())
#define vi vector<int>
#define vvi vector<vector<int>>
#define vc vector<char>
#define vvc vector<vector<char>>
#define vb vector<bool>
#define vvb vector<vector<bool>>
using ll = long long;
using P = pair<int, int>;
/* ----------------よく使う数字や配列----------------- */
int dx[] = {1,0,-1,0};
int dy[] = {0,1,0,-1};
constexpr ll mod = 1e9+7;
constexpr ll inf = INT32_MAX/2;
constexpr ll INF = LLONG_MAX/2;
constexpr long double eps = DBL_EPSILON;
constexpr long double pi = 3.141592653589793238462643383279;
/* ----------------------end----------------------- */

/* --------------------テンプレート------------------ */
template<class T> inline bool chmin(T& a, T b) {
    if (a > b) {
        a = b;
        return true;
    }
    return false;
}
template<class T> inline bool chmax(T& a, T b) {
    if (a < b) {
        a = b;
        return true;
    }
    return false;
}
/* ----------------------end----------------------- */

/* --------------------ライブラリ-------------------- */
ll fact(int i) {       //階乗
    if (i == 0) return 1;
    return (fact(i - 1)) * i % mod;
}
ll gcd(ll a, ll b) {        //最大公約数
    if(b == 0) return a;
    return gcd(b, a % b); 
}
ll lcm(ll a, ll b) {      //最小公倍数
    return a * b / gcd(a, b);
}
int keta(ll n) {        //桁数を求める
    if(n == 0) return 1;
    int count = 0;
    while(n != 0) {
        n /= 10;
        count++;
    }
    return count;
}
ll ketasum(ll n) {    //各桁の和
    ll sum = 0;
    while(n != 0) {
        sum += n % 10;
        n /= 10;
    }
    return sum;
}
/* ----------------------end----------------------- */
#pragma endregion
signed main() {
    int x,k,d;cin >> x >> k >> d;
    int p = x%d;
    // if(p<0)p+=d;
    // cout << p << endl;
    if(abs(x)/d>=k) cout << abs(x)-k*d << endl;
    else {
        k-=abs(x)/d;
        if(k%2==1) cout << min(abs(p-d),abs(p+d)) << endl;
        else cout << abs(p) << endl;
    }
    // cout << -3%5 << endl;
    return 0;
}
/*
-3 2 1
*/