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<cstdlib>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;

typedef long long LL;
LL M,H,CK,Max,Min,ans,a2,a3;
LL a1 = 1 << 30;
int main()
{
	scanf("%I64d %I64d",&H,&M);
	if((H * M)%3 == 0)printf("0\n");
	else
	{
		if(H > M || H == M)
		{
			for(LL i = 1;i < H;i++){
				Max = i*M;
				Min = (H - i) * ((M - (M % 2))/2);
				CK = abs(Max - Min);
				a1 = min(CK,a1);
			}
		}
		else if(H < M)
		{
			for(LL i = 1;i < M;i++){
				Max = i*H;
				Min = (M - i) * ((H - (H % 2))/2);
				CK = abs(Max - Min);
				a1 = min(CK,a1);
			}
		}
		a2 = (H%3) * M;
		a3 = (M%3) * H;
		ans = min(a1,a2);
		ans = min(ans,a3);
		printf("%I64d\n",ans);
	}
}