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)

/*{{{*/
// think twice code once
// Start:[]
// End  :[]
#include <bits/stdc++.h>
using namespace std;

#define rep(i, x, y) for(int i = (x), _ = (y); i <= _; ++ i)
#define per(i, x, y) for(int i = (x), _ = (y); i >= _; -- i)
#define dprintf(...) fprintf(stderr, __VA_ARGS__)
#define disp(x) cout << #x << " = " << x << "; "
#define x first
#define y second
#define mp make_pair
#ifdef __linux__
#define getchar getchar_unlocked
#define LLFORMAT "%lld"
#else
#define LLFORMAT "%I64d"
#endif

typedef long long LL;

template <class T> bool chkmin(T& a, T b) { return a > b ? a = b, true : false; }
template <class T> bool chkmax(T& a, T b) { return a < b ? a = b, true : false; }

template <class T> void read(T& a) {
	char c = getchar(); T f = 1; a = 0;
	for(; !isdigit(c); c = getchar()) if(c == '-') f = -1;
	for(; isdigit(c); c = getchar()) a = a * 10 + c - '0';
	a *= f;
}
/*}}}*/

int N, M;
LL ans;

void work(int N, int M)
{
	rep(i, 1, N)
	{
		LL a = (LL) M * i;
		LL b = (LL) (M / 2) * (N - i);
		LL c = (LL) ((M + 1) / 2) * (N - i);
		if(b == 0 || c == 0) continue;
		//disp(a), disp(b), disp(c), puts("");
		//disp(min(min(abs(a - b), abs(b - c)), abs(a - c))), puts("");
		chkmin(ans, max(max(abs(a - b), abs(b - c)), abs(a - c)));
		b = (LL) ((N - i) / 2) * M;
		c = (LL) ((N - i + 1) / 2) * M;
		chkmin(ans, max(max(abs(a - b), abs(b - c)), abs(a - c)));
	}
}

int main()
{
#ifdef Leeson
	freopen("tmp.in", "r", stdin);
	freopen("tmp.out", "w", stdout);
#endif

	ans = LLONG_MAX;
	read(N), read(M);
	work(N, M);
	work(M, N);
	cout << ans << endl;

	return 0;
}