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 <iostream>
using namespace std;

typedef long long ll;

ll solve(ll h, ll w) {
  ll mini = 1e18;
  for (ll i = 1; i < h - 1; i++) {
    //cout << i * w << endl << (h - i) * (ll)(w / 2.0 + 0.5) << " " << (ll)((h - i) / 2.0 + 0.5) * w << endl;
    ll h_min = min(i * w, min((h - i) * (w / 2), (h - i) * (ll)(w / 2.0 + 0.5)));
    ll h_max = max(i * w, max((h - i) * (w / 2), (h - i) * (ll)(w / 2.0 + 0.5)));
    ll w_min = min(i * w, min(w * ((h - i) / 2), (ll)((h - i) / 2.0 + 0.5) * w));
    ll w_max = max(i * w, max(w * ((h - i) / 2), (ll)((h - i) / 2.0 + 0.5) * w));
    mini = min(mini, min(h_max - h_min, w_max - w_min));
  }

  return mini;
}

int main() {
  ll H, W;
  cin >> H >> W;

  cout << min(solve(H, W), solve(W, H)) << endl;
  return 0;
}