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;

#define ll long long
#define INF 99999999
#define rep(i, n) for(int i = 0; i < (int)(n); i++)
#define all(x) (x).begin(),(x).end()

int main()
{
    ll H,W;
    cin >> H >> W;
    
    ll ans = INF;
    for(ll i = 1;i < H;i++)
    {
        if(W%2==0)
        {
            ll yet = W * (H - i);
            if( abs(yet/2 - W * i) < ans) ans = abs(yet/2 - W * i);
        }
        else
        {
            ll yet1 = (W-1) * (H - i);
            ll yet2 = (W+1) * (H - i);
            ll Max = max(W * i, max( yet1/2, yet2/2));
            ll Min = min(W * i, min( yet1/2, yet2/2));
            if( Max - Min < ans ) ans = Max - Min;
        }

    }
    cout << ans << endl;
}