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;

long long solve(long long x, long long y) {
  long long res = 1e18;
  for (long long i = 1; i < x; ++i) {
    long long a = (x-i)*y;
    long long b = i*(y/2);
    long long c = i*(y-y/2);
    long long mx = max(a, max(b, c));
    long long mn = min(a, min(b, c));
    if (mx - mn < res) res = mx-mn;
  }
  return res;
}

long long solve2(long long x, long long y) {
  long long a = (x/3)*y;
  long long b = (x-2*(x/3))*y;
  return abs(a-b);
}

int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);

  long long h, w;
  cin >> h >> w;

  
  long long a = min(solve(h, w), solve(w, h)); 
  long long b = min(solve2(h, w), solve2(w, h)); 

  cout << min(a, b) << endl;

  return 0;
}