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;
    #define int long long int
    #define loop(i,a,b) for(int i=a;i<b;i++)
    #define scan(arr,n) for (int i = 0; i < n; ++i) cin >> arr[i]
    #define vi vector<int>
    #define si set<int>
    #define pii pair <int, int>
    #define sii set<pii>
    #define vii vector<pii>
    #define mii map <int, int>
    #define pb push_back
    #define faster ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL)
    //Constants
    const int MOD = 1000000007; /* 1e9 + 7*/
    const int MAXN = 1000005; /*1e6 +5 */
     
    int spf[1000009]; // stores the smallest prime factor of a number
     
    void sieve()
    {
    spf[1] = 1;
        for (int i=2; i<MAXN; i++)
            spf[i] = i;
        for (int i=4; i<MAXN; i+=2)
            spf[i] = 2;
     
        for (int i=3; i*i<MAXN; i++)
            if (spf[i] == i)
            {
                for (int j=i*i; j<MAXN; j+=i)
                    if (spf[j]==j)
                        spf[j] = i;
            }
        }
    
     
    /* Iterative Function to calculate (x^y)%p in O(log y) */
    int power(int x,  int y, int p)
    {
        int res = 1;     // Initialize result
        x = x % p; // Update x if it is more than or
                    // equal to p 
        if (x == 0) return 0; // In case x is divisible by p;     
        while (y > 0)
        {
            // If y is odd, multiply x with result
            if (y & 1)
                res = (res*x) % p;     
            // y must be even now
            y = y>>1; // y = y/2
            x = (x*x) % p;
        }
        return res;
    }
    
     
    void solve(int caseNumber)
    {
		// cout<<"Case "<<caseNumber<<": "endl;
		int x,k,d;
		cin>>x>>k>>d;
		int ans=0;
		if(x==k && k==d)
		{
			if(x&1)
			ans=0;
			else
			ans=x;
		}
		else if(k>=x/d)
		{
			int req=x-((x/d)*d);
			int left=k-((x/d)*d);
			if(left&1)
			ans=abs(req-d);
			else
			ans=abs(req);
		}
		else if(k<x/d)
		{
			 ans=abs(x- (k*d));
		}
		cout<<ans<<endl;
			
    }

     
    int32_t main()
    {
        faster;
        /*sieve();*/
       // compute();
      // int t;cin>>t;
     int t=1;
        for(int ii=1;ii<=t;ii++)
        {
            solve(ii);
        }
     
     
        return 0;
    }