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>

#define F first
#define S second
#define pb push_back
#define ll long long
#define int long long
#define ld double

using namespace std ;

const int MAXN = 1001 * 1001 ; 

int a[MAXN] ; 

int32_t main()
{
	ios::sync_with_stdio(0) ; cin.tie(0) ; 
	int n ; 
	cin >> n ; 
	n *= 3 ; 
	for(int i = 0 ; i < n ; i ++ ) cin >> a[i] ; 
	int ans = -1e18 ; 
	multiset<int> s1 , s2 , s3 ; 
	int m = n / 3 ; 
	int sum1 = 0 , sum2 = 0 ; 
	for(int i = 0 ; i < n ; i ++ ) s2.insert(a[i]) , sum2 += a[i] ; 
	while(s2.size()>m)
	{
		sum2 -= *s2.rbegin() ;
		s3.insert(*s2.rbegin()) ; 
		s2.erase(s2.find(*s2.rbegin())) ; 
	}
	for(int i = 0 ; i < 2 * m ; i ++ ) 
	{
		s1.insert(a[i]) ; 
		sum1 += a[i] ; 
	//	cout <<'*'<< sum1 << ' ' << i << endl ; 
		if(s3.find(a[i])!=s3.end()) s3.erase(s3.find(a[i])) ; 
		else s2.erase(s2.find(a[i])) , sum2 -= a[i] ; 	
		while(s1.size()>m)
		{
			sum1 -= *s1.begin() ; 
			s1.erase(s1.find(*s1.begin())) ; 
		}
	//	cout << sum1<< endl ; 
		while(s2.size()<m)
		{
			sum2 += *s3.begin() ; 
			s2.insert(*s3.begin()) ; 
			s3.erase(s3.find(*s3.begin())) ; 
		}
		if(s1.size()>=m)
		{
	//		cout << sum2 << ' ' << sum1 << endl ; 
			ans = max(ans,sum1-sum2) ; 
		}
	}
	cout << ans << endl ; 
}