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: Accepted)

#include <stdio.h>
#include <stdlib.h>
long long int min(long long int x, long long int y){return x<y?x:y;}
long long int max(long long int x, long long int y){return x<y?y:x;}
int main(void){
	long long h,w,i,ans=1e9,s_A,s_B,s_C,ans1,ans2,ans0,rans=1e9;
	scanf("%lld%lld",&h,&w);
	for(i=1;i<h;i++){
		s_A = w*i;
		s_B = (h-i)*(w/2);
		s_C = (h-i)*(w-w/2);
		ans1 = max(max(s_A,s_B),s_C)-min(min(s_A,s_B),s_C);
		s_B = (h-i)/2*w;
		s_C = (h-i-(h-i)/2)*w;
		ans2 = max(max(s_A,s_B),s_C)-min(min(s_A,s_B),s_C);
		ans0 = min(ans1,ans2);
		ans = min(ans,ans0);
	}
	int tmp = h;
	h = w;
	w = tmp;
	for(i=1;i<h;i++){
		s_A = w*i;
		s_B = (h-i)*(w/2);
		s_C = (h-i)*(w-w/2);
		ans1 = max(max(s_A,s_B),s_C)-min(min(s_A,s_B),s_C);
		s_B = (h-i)/2*w;
		s_C = (h-i-(h-i)/2)*w;
		ans2 = max(max(s_A,s_B),s_C)-min(min(s_A,s_B),s_C);
		ans0 = min(ans1,ans2);
		rans = min(rans,ans0);
	}
	ans = min(ans,rans);
	
	
	printf("%lld",ans);
	return 0;
}