Problem:
Score : 400 points

Problem Statement

There is a bar of chocolate with a height of H blocks and a width of W blocks.
Snuke is dividing this bar into exactly three pieces.
He can only cut the bar along borders of blocks, and the shape of each piece must be a rectangle.

Snuke is trying to divide the bar as evenly as possible.
More specifically, he is trying to minimize S_{max} - S_{min}, where S_{max} is the area (the number of blocks contained) of the largest piece, and S_{min} is the area of the smallest piece.
Find the minimum possible value of S_{max} - S_{min}.

Constraints

2 ≤ H, W ≤ 10^5

Input

Input is given from Standard Input in the following format:

H W

Output

Print the minimum possible value of S_{max} - S_{min}.

Sample Input 1

3 5

Sample Output 1

0

In the division below, S_{max} - S_{min} = 5 - 5 = 0.

Sample Input 2

4 5

Sample Output 2

2

In the division below, S_{max} - S_{min} = 8 - 6 = 2.

Sample Input 3

5 5

Sample Output 3

4

In the division below, S_{max} - S_{min} = 10 - 6 = 4.

Sample Input 4

100000 2

Sample Output 4

1

Sample Input 5

100000 100000

Sample Output 5

50000

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<cstdio>
#include<algorithm>
using namespace std;
int main(){
  int h,w,ans;
  scanf("%d%d",&h,&w);
  if(h%3==0 || w%3==0){
    printf("0\n");
    return 0;
  }
  ans=min(h,w);
  int Smax=w*(h/3), Smin=w*(h/3);
  Smax=max(Smax,(h-h/3)*(w/2));
  Smax=max(Smax,(h-h/3)*((w+1)/2));                          
  Smin=min(Smin,(h-h/3)*(w/2));
  Smin=min(Smin,(h-h/3)*((w+1)/2));
  ans=min(ans,Smax-Smin);
  
  Smax=w*(h/3+1), Smin=w*(h/3+1);
  Smax=max(Smax,(h-h/3-1)*(w/2));
  Smax=max(Smax,(h-h/3-1)*((w+1)/2));                          
  Smin=min(Smin,(h-h/3-1)*(w/2));
  Smin=min(Smin,(h-h/3-1)*((w+1)/2));
  ans=min(ans,Smax-Smin);
  
  Smax=h*(w/3), Smin=h*(w/3);
  Smax=max(Smax,(w-w/3)*(h/2));
  Smax=max(Smax,(w-w/3)*((h+1)/2));                          
  Smin=min(Smin,(w-w/3)*(h/2));
  Smin=min(Smin,(w-w/3)*((h+1)/2));
  ans=min(ans,Smax-Smin);
  
  Smax=h*(w/3+1), Smin=h*(w/3+1);
  Smax=max(Smax,(w-w/3-1)*(h/2));
  Smax=max(Smax,(w-w/3-1)*((h+1)/2));                          
  Smin=min(Smin,(w-w/3-1)*(h/2));
  Smin=min(Smin,(w-w/3-1)*((h+1)/2));
  ans=min(ans,Smax-Smin);
  
  printf("%d\n",ans);
  return 0;
}