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.Scanner;

public class Main  {

	public static void main(String[] args) {

/**
 * 		縦に3分割(1<h<w-1)
 *
 *				W
 * 		-----------------------------
 *	H	a	|w -a/2		|
 *			|			|
 *		-----------------------------
 *
 *		縦、横で3分割
 *
 *				W
 * 		-----------------------------
 *		a	|
 *	H		|－－－－－－－－－－－－
 *			|
 *		-----------------------------

 * */

		Scanner sc = new Scanner(System.in);
		long H = sc.nextInt();
		long W = sc.nextInt();
		sc.close();


		long S1 = 0;
		long S2 = 0;
		long S3 = 0;

		long ans = Long.MAX_VALUE;

		// 縦3分割
		for(int a=1; a < W-1; a++) {
			long tmpW = (W-a)/2;
			S1 = H * a;
			S2 = H * tmpW;
			S3 = H * (W - a - tmpW);

			long tmpAns = Math.max(S3,Math.max(S1,S2)) -Math.min(S3, Math.min(S1,S2));

			ans = Math.min(tmpAns, ans);
		}

		// 縦横3分割
		for(int a=1; a < W-1; a++) {
			long tmpH = H/2;
			S1 = H * a;
			S2 = (W-a) * tmpH;
			S3 = (W-a) * (H - tmpH);

			long tmpAns = Math.max(S3,Math.max(S1,S2)) -Math.min(S3, Math.min(S1,S2));

			ans = Math.min(tmpAns, ans);
		}

		///WとＨを入れ替える。同じ値なら入れ替え不要
		if(W==H) {
			System.out.println(ans);
			return;
		}


		for(int a=1; a < H-1; a++) {
			long tmpH = (H-a)/2;
			S1 = W * a;
			S2 = W * tmpH;
			S3 = W * (H - a - tmpH);

			long tmpAns = Math.max(S3,Math.max(S1,S2)) -Math.min(S3, Math.min(S1,S2));

			ans = Math.min(tmpAns, ans);
		}

		for(int a=1; a < H-1; a++) {
			long tmpW = W/2;
			S1 = W * a;
			S2 = (H-a) * tmpW;
			S3 = (H-a) * (W - tmpW);

			long tmpAns = Math.max(S3,Math.max(S1,S2)) -Math.min(S3, Math.min(S1,S2));

			ans = Math.min(tmpAns, ans);
		}

		System.out.println(ans);
	}
}