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)

#include <iostream>
#include <iomanip>
#include <vector>
#include <algorithm>
#include <cstring>
#include <map>
#include <queue>
#include <cmath>
#include <complex> // complex<double> a(1.2 , 2.3);// real(): 1.2, imag()2.3
using namespace std;

#define MOD 1000000007
#define ll long long
#define ld long double
#define FOR(i,a,b) for(ll i=(ll)a;i<(ll)b;i++)
#define rep(i,n) FOR(i,0,n)
#define pb push_back
#define mp make_pair
#define all(a) a.begin(),a.end()
#define rall(a) a.rbegin(),a.rend()
#define rmsame(a) sort(all(a)),a.erase(unique(all(a)), a.end())
#define rmvector(a,b) rep(i,a.size())rep(j,b.size())if(a[i]==b[j]){a.erase(a.begin()+i);i--;break;}
#define pq_pair_tB priority_queue <pair<ll,ll>, vector<pair<ll,ll> > , greater<pair<ll,ll> > > 
#define pq_pair_ts priority_queue <pair<ll,ll> > //第二成分の大きさが関係ない
template<typename X> bool exist(vector<X> vec, X item){return find(all(vec), item)!=vec.end();}
ll gcd(ll a, ll b){if(b==0)return a;return gcd(b,a%b);}
ll lcm(ll a, ll b){return a/gcd(a,b)*b;}

ll cal(ll a,ll b){
    ll ret = 100000000000;
    rep(i, a){
        ll j = b/2;
        ll k = b-j;
        
        ll a1 = i*b;
        ll a2 = j*(a-i);
        ll a3 = k*(a-i);
        
        ll m = min(a1, min(a2, a3));
        ll M = max(a1, max(a2, a3));
        ret = min(ret, M-m);
    }
    return ret;
}

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    
    ll a,b; cin >> a >> b;
    if((a%3 == 0)||(b%3 == 0)) {
        cout << "0" << endl;
        return 0;
    }
    if(a == 1 || b == 1){
        cout << 1 << endl;
        return 0;
    }
    
    ll ans = min(cal(a,b), cal(b,a)) ;
    ans = min(ans, a);
    ans = min(ans, b);
    
    
    
    //cout << fixed << setprecision(16) << ans << endl;
    return 0;
}