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 Ruby (2.3.3) (judge verdict: Accepted)

h,w=gets.split.map &:to_i

def f(a)
	a.max-a.min
end

if h%3==0||w%3==0
p 0
else
p [
f([(h/3)*w,(h/3)*w,(h-h/3-h/3)*w]),
f([(h/3)*w,(h/3+1)*w,(h-h/3-h/3-1)*w]),
f([(w/3)*h,(w/3)*h,(w-w/3-w/3)*h]),
f([(w/3)*h,(w/3+1)*h,(w-w/3-w/3-1)*h]),
f([(h/3)*w,(h-(h/3))*(w/2),(h-(h/3))*(w-w/2)]),
f([(h/3+1)*w,(h-(h/3+1))*(w/2),(h-(h/3))*(w-w/2)]),
f([(w/3)*h,(w-(w/3))*(h/2),(w-(w/3))*(h-h/2)]),
f([(w/3+1)*h,(w-(w/3+1))*(h/2),(w-(w/3+1))*(h-h/2)])
].min
end