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;

template<typename T>
void out(T x) { cout << x << endl; exit(0); }
#define watch(x) cout << (#x) << " is " << (x) << endl




typedef long long ll;
const ll mod = 1e9+7;
const int maxn = 1e6 + 5;











ll div3(ll n, ll m) {
    if (n>m) swap(n, m);


    ll a = n*(m/3);
    ll b = n*((m+2)/3);
    ll res =b-a;
    return res;
}


ll iterate(ll n, ll m) {
    ll res = 1e18;
    for (ll i=1; i<m; i++) {
	vector<ll> v;
	v.push_back(n*i);
	ll rest = m-i;
	v.push_back((n/2)*rest);
	v.push_back(((n+1)/2)*rest);
	sort(v.begin(), v.end());
	res = min(res, v.back()-v.front());
    }
    return res;
}

ll N, M;


int main() {
    ios_base::sync_with_stdio(false); cin.tie(0);  cout.tie(0);

    ll n, m;
    cin>>n>>m;

    N = n;
    M = m;

    if (n%3==0 || m%3==0) out(0);

    if (n>m) swap(n,m);

    vector<ll> v;
    v.push_back(n*(m/2));

    m = (m+1)/2;
    if (n>m) swap(n, m);
    v.push_back(n*(m/2));
    v.push_back(n*((m+1)/2));


    sort(v.begin(), v.end());

    ll ans = v.back()-v.front();


    ans = min(ans, div3(N, M));
    ans = min(ans, iterate(N, M));
    ans = min(ans, iterate(M, N));
    cout<<ans<<endl;

    return 0;
}