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.io.*;
import java.math.*;
import java.text.*;
import java.util.*;
import java.util.regex.*;

public class Main {
  public static void main(String[] args) throws Exception {
    Scanner scanner = new Scanner(System.in);

    int H = scanner.nextInt();
    int W = scanner.nextInt();

    System.out.println(solution(H, W));
    scanner.close();
  }

  private static long solution(int H, int W) {
    if((H%3==0) || (W%3==0)) {
      return 0;
    }

    final long L = Math.max(H, W);
    final long S = Math.min(H, W);

    final long maxArea = S*(L/3+1);
    final long minArea = S*(L/3);

    long min = S;

    for(long i=1;i<S;i++) {
      long j=(long)Math.ceil((double)minArea/i);
      long jMax = Math.min(maxArea/i, L);
      if(0<j && j<jMax) {
        for(;j<jMax;j++) {
          long a = i*j;
          long b = (S-i)*j;
          long c = S*(L-j);
          long thisMin = min(a,b,c);
          long thisMax = max(a,b,c);
          if(thisMax-thisMin<min) min = thisMax-thisMin;
        }
      }
    }
    return min;
  }

  private static long min(long a, long b, long c) {
    return Math.min(a, Math.min(b, c));
  }
  private static long max(long a, long b, long c) {
    return Math.max(a, Math.max(b, c));
  }
}