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;
typedef long long ll;

int main()
{
    ll h, w;
    cin >> h >> w;
    if (h % 3 == 0 || w % 3 == 0)
    {
        cout << 0 << endl;
    }
    else
    {
        ll x = max(h, w);
        ll y = min(h, w);
        ll a = y / 2;
        ll b = y - a;
        ll c = x * y;
        ll ans = x * y;
        ll aaa = 0;
        ll bbb = 0;
        for (int i = 1; i <= x; i++)
        {
            c -= y;
            aaa += a;
            bbb += b;

            ans = min(ans, max({c, aaa, bbb}) - min({c, aaa, bbb}));
        }
        ll k = (x % 3) * y;
        ans = min(ans, k);
        ll d=x/2;
        ll e=x-d;
        c=x*y;
        ll ddd=0;
        ll eee=0;
        for (int i = 1; i <= y; i++)
        {
            c -= x;
            ddd += d;
            eee += e;

            ans = min(ans, max({c, ddd, eee}) - min({c, ddd, eee}));
        }

        ll r=(y%3)*x;
        ans=min(r,ans);
        cout << ans << endl;
    }
}