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: Accepted)

import java.util.*;
import java.awt.*;
import java.awt.geom.*;
import static java.lang.System.*;
import static java.lang.Math.*;
public class Main {
    public static void main(String[] $) {
        Scanner sc = new Scanner(in);
        long[] l = new long[2];
        l[0] = sc.nextLong();
        l[1] = sc.nextLong();
        long ans =  l[0] * l[1];
        long[] s = new long[3];
        for (int i = 0; i < 2; i++) {
            for (int j = 1; j <= l[i]; j++) {
                s[0]=(long)j*l[abs(1-i)];
                s[1]=(l[i]-(long)j)/2*l[abs(1-i)];
                s[2]=l[0]*l[1]-s[0]-s[1];
                Arrays.sort(s);
                ans=min(ans,s[2]-s[0]);
                s[0]=(long)j*l[abs(1-i)];
                s[1]=l[abs(1-i)]/2*(l[i]-(long)j);
                s[2]=l[0]*l[1]-s[0]-s[1];
                Arrays.sort(s);
                ans=min(s[2]-s[0],ans);
            }
        }
        out.println(ans);
    }
    static long sum(long[] a){
        long t=0;
        for (long k:a)t+=k;
        return t;
    }
}