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>
using namespace std ;
#define For(i,a,b) for(int i = (a) , _b = (b) ; i <= _b ; i++ )
#define Ford(i,a,b) for(int i = (a) , _b = (b) ; i >= _b ; i-- )
typedef long long ll ;
typedef pair<int,int> II ;
const int N = 1e6 + 5 ;
const ll base = 1e9 + 7 ;
const ll inf = 1e15 + 7 ;
int n ;
ll a[N] , r[N] , l[N] ;
priority_queue<ll , vector<ll> , greater<ll> > q ;
priority_queue<ll> q1 ;
int main() {
    //freopen(task".inp" , "r" , stdin ) ; freopen(task".out" , "w" , stdout ) ;
    ios_base::sync_with_stdio(false) ; cin.tie(NULL) ;
    cin >> n;
    For(i , 1 , 3*n) cin >> a[i] ;
    /// tinh l[i] ;
    ll to =  0;
    For(i , 1 , n) {
        q.push(a[i]) ;
        to += a[i];
    }
    l[n] = to;
    For(i , n+1 , 3*n) {
        if(q.top() < a[i]) {
            to -= q.top() ;
            q.pop() ;
            q.push(a[i])  ;
            to = to + a[i] ;
        }
        l[i] = to;
    }

    to = 0;
    For(i , n*2+1 , n*3) {
        q1.push(a[i]);
        to+=a[i];
    }
    r[n*2+1] = to;
    Ford(i , n*2 , 1) {
        if(q1.top() > a[i]) {
            to -= q1.top() ;
            q1.pop() ;
            q1.push(a[i]) ;
            to = to + a[i] ;
        }
        r[i] = to;
    }
    ll ans = -inf;
    For(i , n , n*2) ans = max(ans , l[i] - r[i+1]);
    cout << ans ;
    return 0 ;
}