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: Wrong Answer)

#pragma GCC optimize ("O3")
#pragma GCC target ("sse4")
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef long double ld;
typedef vector<int> vi;
typedef pair<int, int> pi;
#define debug(x) cerr << #x << ": " << x << endl;
#define debug2(x, y) debug(x) debug(y);
#define repn(i, a, b) for(int i = (int)(a); i < (int)(b); i++)
#define rep(i, a) for(int i = 0; i < (int)(a); i++)
#define all(v) v.begin(), v.end() 
#define mp make_pair
#define pb push_back
#define lb lower_bound
#define ub upper_bound
#define fi first
#define se second
#define sq(x) ((x) * (x))

template<class T> T gcd(T a, T b){ return ((b == 0) ? a : gcd(b, a % b)); }

int main(){
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	//freopen("input.in", "r", stdin);
	//freopen("output.out", "w", stdout);
	ll n, m;
	cin >> n >> m;
	if(n % 3 == 0 || m % 3 == 0){
		cout << 0 << endl;
		return 0;
	}
	ll ans = 1e18;
	for(ll i = max(1LL, (n / 2) - 100); i <= min(n - 1, (n / 2) + 100); i++){
		debug(i);
		for(ll j = 1; j < m; j++){
			ll a = 1LL * i * j;
			ll b = 1LL * (n - i) * j;
			ll c = 1LL * (m - j) * n;
			ans = min(ans, max({abs(a - b), abs(a - c), abs(b - c)}));
		}
	}
	swap(n, m);
	for(ll i = max(1LL, (n / 2) - 100); i <= min(n - 1, (n / 2) + 100); i++){
		for(ll j = 1; j < m; j++){
			ll a = 1LL * i * j;
			ll b = 1LL * (n - i) * j;
			ll c = 1LL * (m - j) * n;
			ans = min(ans, max({(ll)abs(a - b), (ll)abs(a - c), (ll)abs(b - c)}));
		}
	}
	cout << ans << endl;
	return 0;
}
/*
Things to look out for:
	- Integer overflows
	- Array bounds
	- Special cases
Be careful!
*/