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 (Clang 3.8.0) (judge verdict: Wrong Answer)

#define cmp(N,M) N>M?M:N
#define bigger(N,M) N>M?N:M
#include <stdio.h>
int main(void){
        int n,m,big,small,i,cut,cut2=0,ans=0,ans2=0,ans3=0;
    scanf("%d %d",&n,&m);
    small=cmp(n,m);
    big=bigger(n,m);
    if(small%3==0){
        printf("0\n");
    }else if(small%2==0){
        printf("%d\n",small/2);
    }else{
        for(i=1;i<=big;i++){
            if(small*i*2>(big-i)*small){
                printf("small*i*2=%d (big-i)*small=%d\n",small*i*2,(big-i)*small);
                cut=i-1;
                ans=small*i;
                break;
            }
        }
        for(i=1;i<=small;i++){
            if((big-cut)*i>(small-i)*(big-cut)){
                printf("(big-cut)*i=%d (small-i)*(big-cut)=%d\n",(big-cut)*i,(small-i)*(big-cut));
                cut2=i;
                ans2=(small-i)*(big-cut);
                ans3=(big-cut)*i;
                break;
            }
        }
        printf("cut=%d cut2=%d\n",cut,cut2);
        printf("%d\n",(ans-ans2)>(ans-ans3)?(ans-ans2):(ans-ans3));
        //printf("%d\n",((n-2)/3+1)*n-(n/2+1)*(n-(n-2)/3+1));
    }
    return 0;
}