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 5.4.1) (judge verdict: Wrong Answer)

#include<stdio.h>
#define ll long long
#define min(a,b) ((a)>(b)?(b):(a))
#define max(a,b) ((a)>(b)?(a):(b))
int main(){
  ll h,w;
  scanf("%lld%lld",&h,&w);
  if((h*w)%3==0){
    puts("0");
    return 0;
  }
  ll ans = h*w;
  if(h>w){
    ll t=w;
    w=h;
    h=t;
  }
  //ll x=h/2,y=h/2+h%2;
  for(int i=1;i<w;i++){
    ll x=(w-i)*(h/2),y=(w-i)*(h/2+h%2),z=i*h;
    ans=min((max(y,z)-min(x,z)),ans);
  }
  if(h<w){
    ll t=w;
    w=h;
    h=t;
  }
  ll s,t=1;
  //ll x=h/2,y=h/2+h%2;
  for(int i=1;i<w;i++){
    ll x=(w-i)*(h/2),y=(w-i)*(h/2+h%2),z=i*h;
    ans=min((max(y,z)-min(x,z)),ans);
    if(s!=ans)t=i;
    s=ans;
  }
  ans=min(ans,h);
  printf("%lld\n",ans);
  return 0;
}