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<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long LL;
LL x, y, s1, s2, s3, best = 10000000000;
int main(){
	scanf("%lld%lld", &x, &y);
	if (x*y % 3 == 0) { puts("0"); return 0; }
	if (x%3) best = y; else best = x;
	s1 = x/3*y; s2 = (x-x/3)*(y/2); s3 = x*y-s1-s2;
	best = min(best, max(s1, max(s2, s3)) - min(s1, min(s2, s3)));
	s1 = (x+2)/3*y; s2 = (x-(x+2)/3)*(y/2); s3 = x*y-s1-s2;
	best = min(best, max(s1, max(s2, s3)) - min(s1, min(s2, s3)));
	swap(x, y);
	s1 = x/3*y; s2 = (x-x/3)*(y/2); s3 = x*y-s1-s2;
	best = min(best, max(s1, max(s2, s3)) - min(s1, min(s2, s3)));
	s1 = (x+2)/3*y; s2 = (x-(x+2)/3)*(y/2); s3 = x*y-s1-s2;
	best = min(best, max(s1, max(s2, s3)) - min(s1, min(s2, s3)));
	printf("%lld\n", best);
	return 0;
}