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 <stdio.h>
#include <stdlib.h>
#include <string.h>

#include <iostream>
#include <string>
#include <vector>
#include <map>
#include <unordered_map>
#include <queue>
#include <algorithm>

#include <bits/stdc++.h>
#include <cmath>
#include <limits>

using namespace std;

long long H, W;
vector<long long> v;

int main(int argc, char* argv[]){
	cin.tie(0);
	ios::sync_with_stdio(false);
	
	cin >> H >> W ;
	long long S, s, ans;

	if((H % 3) == 0 || (W % 3) == 0){
		ans = 0;
		printf("%lld\n", ans);
		return 0;
	}

	long long w, h;
	//左1/3、残りの右を上下に分割
	w = W / 3;
	h = H / 2;
	// H*w, h*(W-w), (H-h)*(W-w)
	for(int i=-1; i<=1; i++){
		for(int j=-2; j<=2; j++){
			if(w+i<1 || w+i>=W || h+j<1 || h+j>=H) continue;
			S = max( H * (w+i), max( (h+j)*(W-(w+i)), (H-(h+j))*(W-(w+i)) ) );
			s = min( H * (w+i), min( (h+j)*(W-(w+i)), (H-(h+j))*(W-(w+i)) ) );
			v.push_back(S - s);
		}
	}
	
	//上1/3、残りの下を左右に分割
	h = H / 3;
	w = W / 2;
	// h*W, (H-h)*w, (H-h)*(W-w) 
	for(int i=-2; i<=2; i++){
		for(int j=-1; j<=1; j++){
			if(h+i<1 || h+i>=H || w+j<1 || w+j>=W) continue;
			S = max( (h+i) * W, max( (H-(h+i))*(w+j), (H-(h+i))*(W-(w+j)) ) );
			s = min( (h+i) * W, min( (H-(h+i))*(w+j), (H-(h+i))*(W-(w+j)) ) );
			v.push_back(S - s);
		}
	}

	sort(v.begin(), v.end());
	printf("%lld\n", v[0]);
	return 0;
}