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 Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.util.*;
public class Main {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		long H = sc.nextInt();
		long W = sc.nextInt();
		long ans = Integer.MAX_VALUE;
		if(H % 3 == 0 || W % 3 == 0)
			ans = 0;
		long a, b;
		for(int i = 0; i <= 1; i++) {
			a = (W/3+i)*H;
			b = (W-i-W/3)*(H/2);
			ans = Math.min(ans, Math.abs(a-b));
			if(H % 2 != 0)
				ans = Math.min(ans, W-i-W/3);
			a = (H/3+i)*W;
			b = (H-i-H/3)*(W/2);
			ans = Math.min(ans, Math.abs(a-b));
			if(W % 2 != 0)
				ans = Math.min(ans, H-i-H/3);
		}
		ans = Math.min(ans, Math.min((3-H%3)*W, (3-W%3)*H));
		System.out.println(ans);
	}

}