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 <cstdio>
#include <cstdlib>
#include <cmath>
#include <cctype>
#include <iostream>
#include <sstream>
#include <algorithm>
#include <string>
#include <string.h>
#include <utility>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <stack>

using namespace std;
typedef long long ll;

#ifndef INT_MAX
#  define INT_MAX std::numeric_limits<int>().max()
#endif
#ifndef INT_MIN
#  define INT_MIN std::numeric_limits<int>().min()
#endif
#define LL_MAX std::numeric_limits<ll>().max()
#define LL_MIN std::numeric_limits<ll>().min()
#define INF LL_MAX

#define REP(i,n) for(int i=0;i<n;++i)
#define SORT(name) sort(name.begin(), name.end())

ll H, W;

ll search(ll longer, ll shorter) {
    ll cur_l = longer / 2;
    ll s1 = (shorter / 2) * (longer - cur_l);
    ll s2 = shorter * cur_l;
    ll s3 = W * H - s1 - s2;
    ll min_s = min(min(s1, s2), s3);
    ll max_s = max(max(s1, s2), s3);
    ll ans = abs(max_s - min_s);
    if(ans == 0) {
        return 0;
    }
    ll cur_ans = ans;
    REP(i, 50000) {
        //if(s1 > s2) { cur_l++; }
        //else { cur_l--;}
        cur_l--;
        s1 = (shorter / 2) * (longer - cur_l);
        s2 = shorter * cur_l;
        s3 = W * H - s1 - s2;
        min_s = min(min(s1, s2), s3);
        max_s = max(max(s1, s2), s3);
        cur_ans = abs(max_s - min_s);
        if(cur_ans > ans) { break; }
        ans = cur_ans;
    }
    return ans;
}

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

    if(H % 3 == 0 || W % 3 == 0) {
        printf("0\n");
        return 0;
    }
    if((H % 2 == 0 && W % 2 == 0) ||
       (H % 2 != 0 && W % 2 != 0)) {
        ll longer = (H > W) ? H : W;
        ll shorter = (H < W) ? H : W;
        printf("%lld\n", search(longer, shorter));
        return 0;
    }
    if(H % 2 == 0 || W % 2 ==0) {
        ll even = (H % 2 ==0) ? H : W;
        ll odd = (H % 2 ==0) ? W : H;
        printf("%lld\n", search(odd, even));
        return 0;
    }
    return 0;
}