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: Accepted)

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <queue>
#include <array>
#include <set>
#include <unordered_set>
#include <unordered_map>
#include <cmath>
#include <algorithm>
#include <numeric>
#include <cassert>
#include <utility>
#include <functional>
#include <random>

using namespace std;
using lli = long long int;
template<class T, class U> void init_n(vector<T>& v, size_t n, U x) 
{ v = vector<T>(n, x); }
template<class T> void init_n(vector<T>& v, size_t n) { init_n(v, n, T()); }
template<class T> void read_n(vector<T>& v, size_t n, size_t o = 0)
{ v = vector<T>(n+o); for (lli i=o; i<n+o; ++i) cin >> v[i]; }
template<class T> void read_n(T a[], size_t n, size_t o = 0)
{ for (lli i=o; i<n+o; ++i) cin >> a[i]; }
template<class T> T gabs(const T& x) { return max(x, -x); }
#define abs gabs

lli h, w;

lli score(lli x, lli h, lli w) {
  lli a = (w-x)*h;
  lli b1 = x*(h/2), c1 = x*h-b1, b2 = (x/2)*h, c2 = x*h-b2;
  lli y1 = max({a,b1,c1}) - min({a,b1,c1}),
      y2 = max({a,b2,c2}) - min({a,b2,c2});
  return min(y1, y2);
}

int main() {
  cin >> h >> w;
  lli ans = 1e18;
  for (lli x=1; x<w; ++x) {
    ans = min(ans, score(x, h, w));
  }
  for (lli y=1; y<h; ++y) {
    ans = min(ans, score(y, w, h));
  }
  cout << ans << '\n';
  return 0;
}