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;

typedef long long ll;
typedef vector<ll> vi;
typedef vector<vi> vvi;
typedef vector<pair<ll, ll> > vp;

#define pb push_back
#define rep(i, a, n) for(ll i = (a); i < (n); i++)
#define dep(i, a, n) for(ll i = (a); i >= (n); i--)
#define vin(n, d) rep(i, 0, n) cin >> d[i];
#define out(n) cout << (n) << endl;

__attribute__((constructor))
void initial() {
  cin.tie(0);
  ios::sync_with_stdio(false);
}

int main() {
  ll a, b;
  cin>>a>>b;
  ll ans=min((a%3)*b,(b%3)*a);
  rep(i,1,b)ans=min(ans,max(i*a,max(a/2*(b-i),(a-a/2)*(b-i)))-min(i*a,min(a/2*(b-i),(a-a/2)*(b-i))));
  rep(i,1,a)ans=min(ans,max(i*b,max(b/2*(a-i),(b-b/2)*(a-i)))-min(i*b,min(b/2*(a-i),(b-b/2)*(a-i))));
  // rep(i,1,b)ans=min(ans,max(i*a,(a/2+(a%2==0?0:1))*(b-i))-min(i*a,(a/2)*(b-i)));
  // rep(i,1,a)ans=min(ans,max(i*b,(b/2+(b%2==0?0:1))*(a-i))-min(i*b,(b/2)*(a-i)));
  out(ans)
}