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)

//C - Walking Takahashi
//Hiia
#include <stdio.h>
#include <stdlib.h>//rannsuu, zettaichi(absolute)
#include <time.h>
//srand((unsigned int) time(0));
#include <math.h>//koudona keisann
#include <string.h>//hairetsu
#define max(p,q)((p)>(q)?(p):(q))
#define min(p,q)((p)<(q)?(p):(q))
typedef long long ll;
//#define N 998244353
#define pi 3.1415926535897932384626433832

int gcd(int m,int n){if((0==m)||(0==n))return 0;while(m!=n)if(m>n)m=m-n;else n=n-m;return m;}
int lcm(int m,int n){if((0==m)||(0==n))return 0;return(m*n/gcd(m, n));}
int fact(int x){if(x==1) return(x);else return(x*fact(x-1));}//Kaijou x!
int ruijou(int base,int exponent){int i,answer=1;i=exponent;while(i!=0){answer=answer*base;--i;}return answer;}//ruijou base^exponent

long long int llgcd(long long int m,long long int n){if((0==m)||(0==n))return 0;while(m!=n)if(m>n)m=m-n;else n=n-m;return m;}
long long int lllcm(long long int m,long long int n){if((0==m)||(0==n))return 0;return(m*n/llgcd(m, n));}
long long int llfact(long long int x){if(x==1) return(x);else return(x*llfact(x-1));}//Kaijou
int llruijou(long long int base,long long int exponent){long long int i,answer=1;i=exponent;while(i!=0){answer=answer*base;--i;}return answer;}//ruijou base^exponent

int main()
{
  ll x,k,d;
  scanf("%lld %lld %lld", &x,&k,&d);

  if(x!=0){
    x=max(x,-x);
    if(x-k*d<0){
      if((k-((x/d)+1))%2==0){
        printf("%lld\n", -(x-((x/d)+1)*d));
      }
      else{
        printf("%lld\n", x-((x/d))*d);
      }
    }

    else{
      printf("%lld\n", x-k*d);
    }
  }
  else{
    if(k%2==0){
      printf("0\n");
    }
    else{
        printf("%lld\n", d);
    }
  }

  return 0;
}