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<bits/stdc++.h>
#define mod 1000000007
#define ll long long int
#define ar array
#define mp make_pair
using namespace std;

int main()
{
	ios::sync_with_stdio(false);
	cin.tie(NULL);
	ll h,w;
	cin>>h>>w;
	ll ans = 1e18;
	for(int i = 1; i<=h; i++)
	{
		ll area1 = w*i, area2 = (h-i)/2 * w, area3 = h*w-area1-area2;
		ans = min(ans,max({area1,area2,area3})-min({area1,area2,area3}));
		area2 = w/2 *(h-i), area3 = h*w-area1-area2;
		ans = min(ans,max({area1,area2,area3})-min({area1,area2,area3}));
	}
	for(int i = 1; i<= w;i++)
	{
		ll area1 = h*i, area2 = (w-i)/2 * h, area3 = h*w-area1-area2;
		ans = min(ans,max({area1,area2,area3})-min({area1,area2,area3}));
		area2 = h/2 *(w-i), area3 = h*w-area1-area2;
		ans = min(ans,max({area1,area2,area3})-min({area1,area2,area3}));
	}
	cout<<ans<<endl;
	return 0;
}