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 Java (OpenJDK 11.0.6) (judge verdict: Wrong Answer)

import java.util.*;

public class Main {

    public static void main(String args[]) {

        // 入力
        Scanner sc = new Scanner(System.in);
        long h = Long.parseLong(sc.next());
        long w = Long.parseLong(sc.next());
        sc.close();

        // 主処理
        long result = solve(h, w);

        // 出力
        System.out.println(result);
    }

    private static long solve(long h, long w) {
        if (h % 3 == 0 || w % 3 == 0) {
            return 0;
        }
        long diff = Long.MAX_VALUE;
        PriorityQueue<Long> pq = new PriorityQueue<>();
        for (long i = 1; i < w; i++) { // 最初に縦に切る
            long left = h * i;
            long up = (w - i) * (h / 2);
            long down = (w - i) * (h - h / 2);
            pq.add(left);
            pq.add(up);
            pq.add(down);
            long min = pq.poll();
            pq.poll();
            long max = pq.poll();
            diff = Math.min(diff, max - min);
        }
        for (int i = 1; i < h; i++) { // 最初に横に切る
            long up = w * i;
            long left = (h - i) * (w / 2);
            long right = (h - i) * (w - w / 2);
            pq.add(up);
            pq.add(left);
            pq.add(right);
            long min = pq.poll();
            pq.poll();
            long max = pq.poll();
            diff = Math.min(diff, max - min);
        }
        return diff;
    }
}