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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
using vi = vector<int>;
using vb = vector<bool>;
using vc = vector<char>;
using vs = vector<string>;
using vvi = vector<vector<int>>;
using vvc = vector<vector<char>>;
using pii = pair<int, int>;
#define rep(i,n) for(int i=0; i<(int)(n);i++)

int main(){
  long h,w;
  cin >> h >> w;
  if(h%3 == 0 || w%3 == 0){
    cout << 0 << endl;
    return 0;
  }
  long tmph = h;
  long tmpw = w;
  long x,y,smax,smin;
  long ans1,ans2,ans3,ans4;
  x = tmph / 3;
  if(x == 0) x++;  
  tmph -= x;
  y = tmpw / 2;
  tmpw -= y;
  smax = max(w*x,max(tmph*tmpw,y*tmph));
  smin = min(w*x,min(tmph*tmpw,y*tmph));
  ans1 = smax - smin;
  
  tmph = h;
  tmpw = w;
  x = (tmph / 3) + 1;
  tmph -= x;
  y = tmpw / 2;
  tmpw -= y;
  smax = max(w*x,max(tmph*tmpw,y*tmph));
  smin = min(w*x,min(tmph*tmpw,y*tmph));
  ans2 = smax - smin;
  
  tmph = h;
  tmpw = w;
  x = (tmpw / 3) + 1;
  tmpw -= x;
  y = tmph / 2;
  tmph -= y;
  smax = max(h*x,max(tmph*tmpw,y*tmpw));
  smin = min(h*x,min(tmph*tmpw,y*tmpw));
  ans3 = smax - smin;
  
  tmph = h;
  tmpw = w;
  x = (tmpw / 3);
  if(x == 0) x++;
  tmpw -= x;
  y = tmph / 2;
  tmph -= y;
  smax = max(h*x,max(tmph*tmpw,y*tmpw));
  smin = min(h*x,min(tmph*tmpw,y*tmpw));
  ans4 = smax - smin;
  
  cout << min(min(ans1,ans2),min(ans3,ans4)) << endl;
}