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>
#include <ext/pb_ds/assoc_container.hpp> 
#include <ext/pb_ds/detail/standard_policies.hpp>
#include <ext/rope>

using namespace std;
using namespace __gnu_pbds;
using namespace __gnu_cxx;

#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define endl "\n"
#define long long long 
#define PI 3.141592653589793238

#pragma GCC target ("avx2")
#pragma GCC optimization ("O3")
#pragma GCC optimization ("unroll-loops")

#define TRACE
#ifdef TRACE
#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
    template <typename Arg1>
    void __f(const char* name, Arg1&& arg1){
        cout << name << " : " << arg1 << endl;
        //use cerr if u want to display at the bottom
    }
    template <typename Arg1, typename... Args>
    void __f(const char* names, Arg1&& arg1, Args&&... args){
        const char* comma = strchr(names + 1, ','); cout.write(names, comma - names) << " : " << arg1<<" | ";__f(comma+1, args...);
    }
#else
#define trace(...)
#endif

typedef trie<string,null_type,trie_string_access_traits<>,pat_trie_tag,trie_prefix_search_node_update> pref_trie;
typedef tree<long int,null_type,less<long int>,rb_tree_tag,tree_order_statistics_node_update>ordered_set;
typedef cc_hash_table<long int,long int, hash<long int>, equal_to<long int>, direct_mask_range_hashing<long int>, 
hash_standard_resize_policy<hash_exponential_size_policy<>, hash_load_check_resize_trigger<true>, true>> ht;
typedef gp_hash_table<long int,long int> HashTable;

int main(){
   IOS;
 /*    #ifndef ONLINE_JUDGE
         freopen("in.txt",  "r",  stdin);
         freopen("out.txt", "w", stdout);
     #endif */
   
   long int x,k,d;
   cin>>x>>k>>d;

   x = llabs(x);
   long int ans = 1e15;

   if(x >= k * d){
   	 x -= k * d;
   	 ans = x;
   }
   else{
   		long int minX = x % d;
   		long int stepsGone = x / d;
   		long int stepsLeft = k - stepsGone;

   		if(stepsLeft % 2 == 0)
   			ans = min(ans, minX);
   		else{
   			ans = min(ans,minX + d);
   			ans = min(ans, llabs(minX - d));
   		}

   		if(stepsLeft){
   			minX = d - minX;
   			stepsLeft--;

   			if(stepsLeft % 2 == 0)
   				ans = min(ans,minX);
   			else{
   				ans = min(ans, d - minX);
   				ans = min(ans, minX + d);
   			}
   		}
   }

   cout<<ans;

   return 0;
}