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 <iostream>
#include <algorithm>
#include <vector>
#include <cstdio>
#include <string>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <cmath>
#include <functional>
#include <numeric>
#include <iomanip>
#define fi first
#define se second
#define fcout(n) cout<<fixed<<setprecision((n))
#define cinl(str) getline(cin,(str))
using namespace std;
bool value(int y,int x,int R,int C){return 0<=y&&y<R&&0<=x&&x<C;}
typedef pair<int,int> pii;
typedef long long ll;
typedef vector<int> vi;
typedef vector<long long > vll;
typedef vector< vi > vvi;
double pie=acos(-1);
int INF=1000000007;
int dx[4] = { 0,-1,0,1 };
int dy[4] = { -1,0,1,0 };


int main() {
	long long S,s1,s2,s3,ans,x,y,z;
	double h,w;
	long long H,W;
	long long a,b;
	cin>>h>>w;
	H=h;
	W=w;
	S = h*w;
	if(S%3==0){
		cout<<0<<endl;
		return 0;
	}

	if(h>=w){
		a = 2*h/3.0 + 0.5;
		s1 = (H-a)*W;
		s2 = max(a*(W/2),(a/2)*W);
		s3 = S - s1 - s2;
	}
	else if(H<W){
		b = 2*w/3.0 + 0.5;
		s1 = (W-b)*H;
		s2 = max(b*(H/2),(b/2)*H);
		s3 = S - s1 - s2;
	}
	x = abs(s1-s2);
	y = abs(s2-s3);
	z = abs(s3-s1);
	ans = max(x,y);
	ans = max(ans,z);
	//cout<<s1<<" "<<s2<<" "<<s3<<endl;
	cout<<ans<<endl;
}