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)

//bad_bat
#include<bits/stdc++.h>
using namespace std;

#define Fast           ios_base::sync_with_stdio(0);cin.tie(0),cout.tie(0)
#define FO             freopen("in.txt", "r", stdin)
#define FC             freopen("out.txt", "w", stdout)
#define aise           cout<<"aise"<<endl
#define pb             push_back
#define pf             push_front
#define ppb            pop_back
#define ppf            pop_front
#define mp             make_pair
#define ll             long long
#define all(x)         x.begin(), x.end()
#define mset(v, a)     memset(v, a, sizeof(v))
#define pll            pair< ll, ll >
#define pd             pair< double, double >
#define ff             first
#define ss             second
#define pi             acos(-1.0)
#define mxN            2010
#define inf            1e14
#define mod            1000000007
#define tc             ll tc;scanf("%lld",&tc);while(tc--)

//NCR iterative
/*voidcomb_table(intN)
{
    for(inti=0; i<=N; ++i)
    {
        for(intj=0; j<=i; ++j)
        {
            if(j==0or j==i)
            {
                C[i][j]=1LL;
            }
            else
            {
                C[i][j]=(C[i-1][j-1]+C[i-1][j]);
            }
        }
    }
}*/


//NCR recursive
/*ll dp[55][55];

ll NCR(ll n,ll r)
{

    if(n==r)
        return dp[n][r]=1;
    if(r==0)
        return dp[n][r]= 1;
    if(r==1)
        return dp[n][r]=n ;
    if(dp[n][r] != -1)
        return dp[n][r];
    return dp[n][r] = NCR(n-1,r)+NCR(n-1,r-1);
}*/

//Sieve:
/*bool Flag[1000010];
vector<ll> Prime;

void sieve()
{

    ll val=sqrt(1000010);
    for(ll i=4;i<=1000005; i+=2) Flag[i]=1;
    for(ll i=3;i <= val;i+=2)
    {

        if(Flag[i]==0)
        {

            for(ll j=i*i;j<=1000005;j+=i) Flag[j]=1;
        }
    }
    Flag[1]=1;
    for(ll i=1;i<=1000000;i++)
        if(Flag[i]==0) Prime.pb(i);
}*/


//Conversion
/*string bin(ll x)
{

    string ret;
    while(x)
    {

        ll t=x%2;
        char c;
        if(t==1)
            c='1';
        else
            c='0';
        ret.pb(c);
        x/=2;
    }
    reverse(all(ret));
    return ret;
}*/

/*ll lcm(ll x, ll y)
{

    ll ret=__gcd(x,y);
    return (x*y)/ret;
}*/

/*ll dx[]= {0,0,1,-1};
ll dy[]= {1,-1,0,0};*/

int main()
{

    ll x,k,d;
    cin >> x >> k >> d;
    if(x < 0) x*=-1;
    ll vf=x/d;
    if(x%d) vf++;
    if(k<=vf)
    {

        ll ans=x-k*d;
        cout << abs(ans);
        return 0;
    }
    k-=vf;
    ll a=x-vf*d;
    ll b=a+d;
    if(k%2) cout << abs(b);
    else cout << abs(a);
}