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 <bits/stdc++.h>
#define FOR(i,a,b) for(int i=(a);i<(b);i++)
#define REP(i,n) for(int i=0;i < (n);i++)
#define ALL(obj) (obj).begin(),(obj).end()
#define SCANF(i) int i;scanf("%d",&i)
typedef long long ll;
using namespace std;
using P = pair<int,int>;
const long long INF = 1LL << 60;
const int IINF=100000000;
const int MOD = (int)1e9 + 7;
//約数列挙
vector<long long> divisor(long long n) {
    vector<long long> ret;
    for (long long i = 1; i * i <= n; i++) {
        if (n % i == 0) {
            ret.push_back(i);
            if (i * i != n) ret.push_back(n / i);
        }
    }
    //sort(ret.begin(), ret.end()); // 昇順に並べる
    return ret;
}
vector<int> dx={1,0,-1,0};vector<int> dy={0,-1,0,1};

signed main () {
    ll x,k,d;cin >> x >> k >> d;
    x = abs(x);
    ll ans;
    if (x/d >= k) {
        ans = abs(x - d*k);
    } else {
        if((k - x/d)%2==0)ans=x%d;
        else ans=abs(x%d-d);
    }
    cout << ans << endl;
}//https://atcoder.jp/contests/abc175/tasks/abc175_c