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;

#define FOR(i,a,b) for(int i=(a);i<(b);i++)
#define REP(i,n) FOR(i,0,n)
#define ALL(v) (v).begin(),(v).end()

template<typename A, typename B> inline bool chmax(A &a, B b) { if (a<b) { a=b; return 1; } return 0; }
template<typename A, typename B> inline bool chmin(A &a, B b) { if (a>b) { a=b; return 1; } return 0; }

typedef unsigned long long ull;
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef pair<int, pii> P;

const ll INF = 1ll<<29;
const ll MOD = 1000000007;
const double EPS  = 1e-10;

ll h, w;

ll solve() {
	ll res = 1ll<<62;
	
	for (ll i = 1; i < h; i++) {
		ll s[3];
		s[0] = i * w;
		s[1] = (h - i) * (w / 2);
		s[2] = (h - i) * ((w + 1) / 2);
		
		sort(s, s + 3);
		chmin(res, s[2] - s[0]);
	}
	
	for (ll i = 1; i < h - 1; i++) {
		ll s[3];
		s[0] = i * w;
		s[1] = (h - i) / 2 * w;
		s[2] = (h - i + 1) / 2 * w;
		
		sort(s, s + 3);
		chmin(res, s[2] - s[0]);
	}
	
	return res;
}

int main() {
	cin >> h >> w;
	
	ll ans = 1ll<<60;
	
	chmin(ans, solve());
	swap(h, w);
	chmin(ans, solve());
	
	cout << ans << endl;
	
	return 0;
}