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 <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>

#define INF 1e9+7
#define pf2 pop_front
#define pb2 pop_back
#define pb push_back
#define pf push_front
#define fi first
#define se second
#define sz size
#define eps 1e-7
#define fod find_by_order
#define fastio ios::sync_with_stdio(0);cin.tie(NULL);cout.tie(NULL);
#define ofk order_of_key
#define val(x) cout << "Value dari "<< #x << " adalah " << x  << "\n"
#define tr tree<int,null_type,less<int>,rb_tree_tag,tree_order_statistics_node_update>
typedef long long ll;
using namespace __gnu_pbds;
using namespace std;

int dx[8] = {1,0,-1,0,1,1,-1,-1};
int dy[8] = {0,1,0,-1,1,-1,1,-1};

void readf(string x){
	freopen((x+".in").c_str(),"r",stdin);
	freopen((x+".out").c_str(),"w",stdout);
}


int read()
{
	bool min = 0;
	int  result = 0;
	char ch;
	ch = getchar();
	while(1)
	{
		if(ch == '-') break;
		if(ch >='0' && ch <= '9') break;
		ch = getchar();
	}
	if(ch == '-') min = 1;else result = ch-'0';
	while(1)
	{
		ch =getchar();
		if(ch< '0' || ch>'9') break;
		result = result * 10 + (ch-'0');
	}
	if(min) return -result;
	return result;
}
//Reynaldo's Template

int main(){
	long long N,M;
	cin >> N >> M;
	ll has = N * M;
	if(N % 3 == 0 || M % 3 == 0){
		cout << 0 << "\n";
		return 0;
	}
	else if(N % 3 == 1){
		ll mini = 1e18;
		for(int i=1;i<=N-1;i++){
			mini = min(mini,abs(i*M - (has - i*M)/2));
		}
		cout << mini << "\n";
	}
	else if (M % 3 == 1){
		ll mini = 1e18;
		for(int i=1;i<=M-1;i++){
			mini = min(mini,abs(i*N - (has - i*N)/2));
		}
		cout << mini << "\n";
	}
	else{
		ll mini =1e18;
		for(int i=1;i<=N-1;i++){
			mini = min(mini,abs(max(i*M,max((N-i)*(N/2 + 1),(N-i) * N/2))) - min(i*M,min((N-i)*(N/2 + 1),(N-i)*(N/2 ))));
		}
		cout << mini << "\n";
	}
}