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;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll, ll> PA;
typedef priority_queue<int, vector<int>, greater<int> > PQG;
#define REP(i, a, n) for(ll i = (ll) a;i < (ll) n;i++)
#define fi first
#define se second
#define pb push_back
#define INF 1000000009
#define INFL 1000000000000000018
#define MOD 1000000007
#define MAXN 200005
#define MAXM 200005

ll h, w, ans = INFL;

ll mx(ll a, ll b){
    return a > b ? a : b;
}

ll mn(ll a, ll b){
    return a < b ? a : b;
}

int main()
{
    cin >> h >> w;
    if((h * w) % 3 == 0){
        cout << 0 << endl;
        return 0;
    }
    REP(i, 1, h){
        ll a = i * w, b = (h - i) * ((w + 1) / 2), c = (h - i) * (w / 2);
        ll ma = mx(a, mx(b, c));
        ll mi = mn(a, mn(b, c));
        ans = mn(ans, (ma - mi));
        //printf("ans %lld\n", ans);
    }
    REP(i, 1, w){
        ll a = i * h, b = (w - i) * ((h + 1) / 2), c = (w - i) * (h / 2);
        ll ma = mx(a, mx(b, c));
        ll mi = mn(a, mn(b, c));
        ans = mn(ans, (ma - mi));
        //printf("ans %lld\n", ans);
    }
    if(w != 2 && h != 2){
        ll a = ((w + 2) / 3) * h - (w / 3) * h;
        ans = mn(ans, a);
        //printf("ans %lld\n", ans);
        ll b = ((h + 2) / 3) * w - (h / 3) * w;
        ans = mn(ans, b);
        //printf("ans %lld\n", ans);
    }
    cout << ans << endl;


    return 0;
}
/*
*/