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>
using namespace std;

#ifdef _DEBUG
#define dout cout
#define debug() if (true)
#define dvout(v) vout(v)
#else
#define dout if (false) cout
#define debug() if (false)
#define dvout(v) if(false) vout(v)
#endif

#define rep(i, n) for (int i = 0, i##_len = (n); i < i##_len; ++i)
#define all(x) (x).begin(), (x).end()
#define vout(v)           \
	for (auto &i : v)     \
	{                     \
		cout << i << " "; \
	}                     \
	cout << "\n"; //配列出力
#define vec(v, n) vector<int> v(n, 0);
#define vecll(v, n) vector<ll> v(n, 0);
#define vecvec(v,n,m) vector<vector<int>> v(n, vector<int>(m));
#define vin(v)        \
	for (auto &i : v) \
		cin >> i; //配列入力

#define mod (int)(1e9+7)
typedef long long ll;
typedef unsigned long long ull;
#define next asdnext
#define prev asdprev
#define _n "\n"
#define _n2 "\n\n"
#define _t "\t"

//	LLONG_MAX 9.2*1e18	9223372036854775807LL = (2^63)-1
//	INT_MAX 2.1*1e9 2147483647 = (2^31)-1
//	O(1e8) 1.2sec

void Main()
{
	ll X,K,D;
	cin>>X>>K>>D;
	ull min = abs(X);
	if(K<=abs(X/D)){
		dout<<"flag"<<'1'<<_n;
		if(X>0) cout<<abs(X-D*K)<<_n;
		else cout<<abs(X+D*K)<<_n;
		return;
	}

	else if(((X/D)%2)==(K%2)){
		dout<<"flag"<<'2'<<_n;
		cout<<abs(X%D)<<_n;
		return;
	}

	else{
		dout<<"flag"<<'3'<<_n;
		if(X%D<0)cout<<abs(X%D+D)<<_n;
		else cout<<abs(X%D-D)<<_n;
		return;
	}

}


void Test();
int main()
{
	cin.tie(nullptr);
	ios_base::sync_with_stdio(false);
	cout << std::fixed << std::setprecision(15);
	Main();
	return 0;
}