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>
typedef  long long int ll;
typedef  long double ld;
#define sync ios_base::sync_with_stdio(false); cin.tie(NULL) 
#define input(arr,n) for(ll i1=0;i1<n;i1++ )cin>>arr[i1]
#define mod 1000000007
#define F first
#define S second 
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#pragma GCC optimize("O3")
//recursions\
#pragma comment(linker, "/stack:200000000")
//loops\
#pragma GCC optimize("unroll-loops")
using namespace __gnu_pbds;
#define ordered_set tree<ll, null_type,less_equal<ll>, rb_tree_tag,tree_order_statistics_node_update>//s.order_of_key(val) *s.find_by_order(ind)
using namespace std;
const int N = 1e6+5;
int main()
{
	sync;
	ll x, k, d;
	cin >> x >> k >> d;
	ll ans = abs(x);
	if(x == 0)
		return cout << 0, 0;
	ll minm = 1e16 / d + 1;
	if(k % 2 != minm % 2) minm++;
	k = min(k, minm);
	if(x < 0)
		d = abs(d);
	if(x > 0)
		d = -abs(d);
	ll low = k & 1, high = (k + 1) / 2;
	while(low <= high)
	{
		ll mid = (low + high) / 2, pt = 2 * mid;
		if(k & 1)
			pt--;
		ll pos = x + pt * d;
		if(x < 0 && pos <= 0)
		{
			ans = min(ans, abs(pos));
			low = mid + 1;
		}
		else if(x > 0 && pos >= 0)
		{
			
			ans = min(ans, abs(pos));
			low = mid + 1;
		}
		else
			high = mid - 1;
			
	}
	low = k & 1, high = (k + 1) / 2;
	while(low <= high)
	{
		ll mid = (low + high) / 2, pt = 2 * mid;
		if(k & 1)
			pt--;
		ll pos = x + pt * d;
		if(x < 0 && pos >= 0)
		{
			ans = min(ans, abs(pos));
			high = mid - 1;
		}
		else if(x > 0 && pos <= 0)
		{
			
			ans = min(ans, abs(pos));
			high = mid - 1;
		}
		else
			low = mid + 1;
			
	}
	cout << ans;
}