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 <algorithm>
#include <cstdio>
#include <iostream>
#include <map>
#include <math.h>
#include <queue>
#include <set>
#include <stack>
#include <vector>
using namespace std;
typedef long long int ll;
typedef long double ld;
#define rep(i,n) for(int i = 0; i < (n); ++i)
#define Sort(Array) sort(Array.begin(), Array.end())
#define Reverse(a) reverse(a.begin(), a.end())
#define out(ans) cout << ans << endl;
const int MOD = 1000000007;
const int INF = 2147483647;
const ld PI = 3.14159265358979323846;

//------------↓------------- M -------------- E ---------------- M --------------- O ---------------↓--------------//
// コンパイル 
// g++ -std=c++1z
//
// -------型変換--------
// int を string に変換
// string s = to_string(n);
// string を int に変換
// int n = stoi(s);
//
// -------二分探索---------
// k以上の値が最初に現れる時のイテレータ
// lower_bound(data.begin(), data.end(), k)
// kより大きい値が最初の現れる時のイテレータ O(logN)
// upper_bound(data.begin(), data.end(), k)
// kがdataに存在するかをtrue or falseで返す O(logN)
// binary_search(data.begin(), data.end(), k)
// 
//
//
//
//
//
// 
//------------↑------------- M -------------- E ---------------- M --------------- O ---------------↑--------------//



int main() {

    ll X, K, D; cin >> X >> K >> D;

    ll ans;
    if (log(X) >= log(K) + log(D)) {
        ans = X - K*D;
    } else {
        ll pls = X%D;
        ll mns = X%D - D;
        if (X/D % 2 == 0) {
            if (K % 2 == 0) {
                ans = pls;
            } else {
                ans = mns;
            }
        } else {
            if (K % 2 == 1) {
                ans = pls;
            } else {
                ans = mns;
            }
        }
    }
    ans = abs(ans);

    cout << ans << endl;


    return 0;
}