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>
#include <stdlib.h>
typedef long long LL;
#define Rep(i,a,n)for(int i=(int)(a);i<(int)(n);i++)
#define rep(i,n)Rep(i,0,n)
#define Ma(x,y)((x)>(y)?(x):(y))
#define mi(x,y)((x)<(y)?(x):(y))
int upLL(const void*a,const void*b){return*(LL*)a-*(LL*)b;}

int main(){
  LL a[2],ans=1e11;
  scanf("%lld%lld",a,a+1);
  if(!(a[0]%3*a[1]%3)){printf("0");return 0;}
  rep(i,2){//たてよこ対等
    LL x=a[i],y=a[i?0:1],z[3],s=x*y;
    Rep(j,Ma(1,x/3-2),mi(x,x/3+3)){//||,|-
      z[0]=y*j,z[1]=y*((x-j)/2),z[2]=s-z[0]-z[1];
      qsort(z,3,8,upLL);
      ans=mi(ans,z[2]-z[0]);
      z[0]=y*j,z[1]=(y/2)*(x-j),z[2]=s-z[0]-z[1];
      qsort(z,3,8,upLL);
      ans=mi(ans,z[2]-z[0]);
    }
  }
  printf("%lld",ans);
  return 0;
}