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 <iostream>
#include <fstream>
#include <cstdio>
#include <cmath>
#include <vector>
#include <string>
#include <set>
#include <map>
#include <stack>
#include <queue>
#include <deque>
#include <bitset>
#include <algorithm>
#include <complex>
#include <array>
#include <iomanip>
using namespace std;

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

typedef long long ll;
typedef vector<int> VI;
typedef vector<ll> VL;
typedef vector<long double> VD;
typedef vector<VI> VVI;
typedef vector<VL> VVL;
typedef vector<VD> VVD;
typedef pair<int,int> P;
typedef pair<ll,ll> PL;

template<typename T> void chmin(T &a, T b) { if (a > b) a = b; }
template<typename T> void chmax(T &a, T b) { if (a < b) a = b; }

int in() { int x; scanf("%d", &x); return x; }
ll lin() { ll x; scanf("%lld", &x); return x; }
#define INF 1LL<<60

int main() {
    ll H, W;
    cin >> H >> W;

    // 縦
    ll w_opt = 1e18;
    FOR(w, 1, W-1) {
        ll S1 = H * w;
        ll S2_tmp1, S2_tmp2, S3_tmp1, S3_tmp2;
        S2_tmp1 = H * ((W-w) / 2), S3_tmp1 = H * ((W-w+1) / 2);
        S2_tmp2 = (W-w) * (H/2), S3_tmp2 = (W-w) * ((H+1) / 2);

        ll cand1 = max(S1, max(S2_tmp1, S3_tmp1)) - min(S1, min(S2_tmp1, S3_tmp1));
        ll cand2 = max(S1, max(S2_tmp2, S3_tmp2)) - min(S1, min(S2_tmp2, S3_tmp2));
        w_opt = min(w_opt, min(cand1, cand2));
    }

    // 横
    ll h_opt = 1e18;
    FOR(h, 1, H-1) {
        ll S1 = h * W;
        ll S2_tmp1, S2_tmp2, S3_tmp1, S3_tmp2;
        S2_tmp1 = W * ((H-h) / 2), S3_tmp1 = W * ((H-h+1) / 2);
        S2_tmp2 = (H-h)* (W/2), S3_tmp2 = (H-h) * ((W+1) / 2);

        ll cand1 = max(S1, max(S2_tmp1, S3_tmp1)) - min(S1, min(S2_tmp1, S3_tmp1));
        ll cand2 = max(S1, max(S2_tmp2, S3_tmp2)) - min(S1, min(S2_tmp2, S3_tmp2));
        h_opt = min(h_opt, min(cand1, cand2));
    }

    cout << min(w_opt, h_opt) << endl;

    return 0;
}