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<cstdio>
#define ll long long int
ll ma(ll n)
{
    if (n%2==0)return n/2;
    else return n/2+1;
}
ll maxn(ll a,ll b)
{
    return a>b?a:b;
}
ll minn(ll a,ll b)
{
    return a<b?a:b;
}
int main()
{
    ll H,M;
    while (scanf("%lld %lld",&H,&M)==2)
    {
        ll cur;
        if (H%3==0||M%3==0)printf("0\n");
        else
        {
           // cur=10000000000;
            cur=maxn(maxn(1*M,(H-1)*(M/2)),(H-1)*((M+1)/2))-minn(minn(1*M,(H-1)*(M/2)),(H-1)*((M+1)/2));//@@@@1
            for (ll i=2;i<H;i++)
            {
                cur=minn(maxn(maxn(i*M,(H-i)*(M/2)),(H-i)*((M+1)/2))-minn(minn(i*M,(H-i)*(M/2)),(H-i)*((M+1)/2)),cur);
            }
            ll t=H;H=M;M=t;
         //   cur=minn(maxn(maxn(1*M,(H-1)*(M/2)),(H-1)*ma(M))-minn(minn(1*M,(H-1)*(M/2)),(H-1)*ma(M)),cur);
            for (ll i=1;i<H;i++)
            {
                cur=minn(maxn(maxn(i*M,(H-i)*(M/2)),(H-i)*((M+1)/2))-minn(minn(i*M,(H-i)*(M/2)),(H-i)*((M+1)/2)),cur);
            }
            printf("%lld\n",cur);
        }
    }
    return 0;
}