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 Rust (1.15.1) (judge verdict: Wrong Answer)

use std::cmp::*;
use std::io::*;

fn main() {
    let mut s: String = String::new();
    std::io::stdin().read_to_string(&mut s).ok();
    let mut itr = s.trim().split_whitespace();
    let h: i64 = itr.next().unwrap().parse().unwrap();
    let w: i64 = itr.next().unwrap().parse().unwrap();

    let mut ans = 1i64 << 60;
    for i in 0..h / 2 {
        let a = w * (i + 1);
        let rem = h - (i + 1);
        // tate
        let mut b = (w / 2) * rem;
        let mut c = ((w + 1) / 2) * rem;
        ans = min(ans, max(a, max(b, c)) - min(a, min(b, c)));
        // 残ったものを半分にする。
        // yoko
        b = (rem / 2) * w;
        c = ((rem + 1) / 2) * w;
        ans = min(ans, max(a, max(b, c)) - min(a, min(b, c)));
    }
    for i in 0..w / 2 {
        let a = h * (i + 1);
        let rem = w - (i + 1);
        // tate
        let mut b = (h / 2) * rem;
        let mut c = ((h + 1) / 2) * rem;
        ans = min(ans, max(a, max(b, c)) - min(a, min(b, c)));
        // 残ったものを半分にする。
        // yoko
        b = (rem / 2) * h;
        c = ((rem + 1) / 2) * h;
        ans = min(ans, max(a, max(b, c) - min(a, min(b, c))));
    }
    println!("{}", ans);
}