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)

#pragma region include
#include <iostream>
#include <vector>
#include <algorithm>
#include <map>
#include <string>
#include <queue>
#include <stack>
#include <cmath>
#include <set>
#include <cstdio>
#define ALL(obj) (obj).begin(),(obj).end()
#define RALL(obj) (obj).rbegin(),(obj).rend()
#define REP(i, n) for(int i = 0; i < n; i++)
#define REPR(i, n) for(int i = n; i >= 0; i--)
#define FOR(i, m, n) for(int i = m; i < n; i++)
#define MOD 1000000007
#define INF 1000000000
#define LLINF 4000000000000000000
using namespace std;
typedef long long ll;
typedef pair<int, int> PII;
typedef vector<int> VI;
typedef vector<VI> VVI;
#pragma endregion

//#define __DEBUG__
#ifdef  __DEBUG__
#define dump(x) cerr << #x << " = " << (x) << " [" << __LINE__ << ":" << __FUNCTION__ << "] " << endl;
// vector出力
template<typename T>
ostream& operator << (ostream& os, vector<T>& v) {
    os << "{";
    REP(i, (int)v.size()) { os << v[i] << (i < v.size() - 1 ? ", " : ""); }
    os << "}";
    return os;
}
// pair出力
template<typename T, typename U>
ostream& operator << (ostream& os, pair<T, U>& p) {
    return os << "(" << p.first << ", " << p.second << ")";
}
// map出力
template<typename T, typename U>
ostream& operator << (ostream& os, map<T, U>& map_var) {
    os << "{";
    for (auto itr = map_var.begin(); itr != map_var.end(); itr++) {
        os << "(" << itr->first << ", " << itr->second << ")";
        itr++;
        if (itr != map_var.end()) os << ", ";
        itr--;
    }
    os << "}";
    return os;
}
// set 出力
template<typename T>
ostream& operator << (ostream& os, set<T>& set_var) {
    os << "{";
    for (auto itr = set_var.begin(); itr != set_var.end(); itr++) {
        os << *itr;
        ++itr;
        if (itr != set_var.end()) os << ", ";
        itr--;
    }
    os << "}";
    return os;
}
#endif

int main() {
    cin.tie(0); ios::sync_with_stdio(false);
    int H, W; cin >> H >> W;
    int ans = INF;
    if (H % 3 == 0 || W % 3 == 0) {
        cout << 0 << endl;
    }
    else {
        REP(i, H) {
            ans = min(ans, max({i * W, W / 2 * (H - i), H * W - W / 2 * (H - i) - i * W}) - min({i * W, W / 2 * (H - i), H * W - W / 2 * (H - i) - i * W}));
        }
        REP(i, W) {
            ans = min(ans, max({i * H, H / 2 * (W - i), H * W - H / 2 * (W - i) - i * H}) - min({i * H, H / 2 * (W - i), H * W - H / 2 * (W - i) - i * H}));
        }
        cout << ans << endl;
    }
    getchar(); getchar();
}