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 C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>

using namespace std;

const int maxn = 1e5 + 100;
int s[500];
int t[50];

int main() {
    long long n, m;
    long long a, b;

    cin >> n >> m;
    if (n % 3 == 0 || m % 3 == 0) {
        cout << 0 << endl;
    } else {
        if (n % 2 == 0 || m % 2 == 0) {
            if (n % 2 == 0) {
                a = n;
                b = m;
            } else {
                a = m;
                b = n;
            }
            if (n % 2 == 0 && m % 2 == 0) {
                a = min(n, m);
                b = max(n, m);
            }
            long long temp = a / 2;
            long long ans = 999999999999999999;
            for (int i = 1; i < b; i++) {
                ans = min(ans, abs(i * temp - (b - i) * a));
            }
            ans = min(ans,min(m,n));
            cout << ans << endl;

        } else {
            a = m + n;
            b = max(m, n);
            a = a - b;
            long long temp = a / 2;
            long long ans = 999999999999999999;
            for (int i = 1; i < b; i++) {
                long long aa = i * temp;
                long long bb = i * (temp + 1);
                long long cc = a * (b - i);
                ans = min(ans, max(aa, max(bb, cc)) - min(aa, min(bb, cc)));
            }
            ans = min(ans,min(m,n));

            cout << ans << endl;
        }
    }
    return 0;
}