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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>

typedef unsigned	us;

us max(us a, us b)
{
	if(a > b)
		return a;
	else
		return b;
}

us min(us a, us b)
{
	if(a < b)
		return a;
	else
		return b;
}

int main(void)
{
	us high, wide;
	
	scanf("%u %u", &high, &wide);
	
	if( (high%3 == 0) || (wide%3 == 0) )
		printf("0\n");
	else{
		us s_max = 0;
		us s_min = ~0;
		us ans = ~0;
		us i;
		
		for(i=1; i<high; i++){
			us a, b, c;
			us h = (high-i)/2;
			us w = wide/2;
			
			a = i*wide;
			b = (high-i)*w;
			c = (high-i)*(wide-w);
			s_max = max(a, max(b, c));
			s_min = min(a, min(b, c));
			
			ans = min(ans, s_max-s_min);
			
			b = h*wide;
			c = ((high-i)-h)*wide;
			s_max = max(a, max(b, c));
			s_min = min(a, min(b, c));
			
			ans = min(ans, s_max-s_min);
			
			//printf("high:%d max:%d min:%d\n", i, s_max, s_min);
			//printf("high:%d a:%d b:%d c:%d\n", i, a, b, c);
		}
		for(i=1; i<wide; i++){
			us a, b, c;
			us h = high/2;
			us w = (wide-i)/2;
			
			a = i*high;
			b = (wide-i)*h;
			c = (wide-i)*(high-h);
			s_max = max(a, max(b, c));
			s_min = min(a, min(b, c));
			
			ans = min(ans, s_max-s_min);
			
			b = w*high;
			c = ((wide-i)-w)*high;
			s_max = max(a, max(b, c));
			s_min = min(a, min(b, c));
			
			ans = min(ans, s_max-s_min);
			
			//printf("wide:%d max:%d min:%d\n", i, s_max, s_min);
			//printf("wide:%d a:%d b:%d c:%d\n", i, a, b, c);
		}
		
		printf("%d\n", ans);
	}
	
	
	return 0;
}