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 (GCC 9.2.1) (judge verdict: Accepted)

#include<stdio.h>
int main(){
  long long h,w,i,min;
  scanf("%lld%lld",&h,&w);
  min=h*w;
  if(h>=3){
    if(h%3==0)
      min=0;
    else
      min=w;
  }
  if(w>=3){
    if(w%3==0)
      min=0;
    else if(min>h)
      min=h;
  }
  for(i=1;i<=w/2;i++){
    long long a,b,c;
    a=i*h;
    b=(w-i)*(h/2+h%2);
    c=(w-i)*(h/2);
    if(a<c){
      if(b-a<min)
        min=b-a;
    }
    else if(b<a){
      if(a-c<min)
        min=a-c;
    }
    else{
      if(b-c<min)
        min=b-c;
    }
  }
  for(i=1;i<=h/2;i++){
    long long a,b,c;
    a=i*w;
    b=(h-i)*(w/2+w%2);
    c=(h-i)*(w/2);
    if(a<c){
      if(b-a<min)
        min=b-a;
    }
    else if(b<a){
      if(a-c<min)
        min=a-c;
    }
    else{
      if(b-c<min)
        min=b-c;
    }
  }
  printf("%lld\n",min);
  return 0;
}