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 (GCC 5.4.1) (judge verdict: Accepted)

#include<stdio.h>
#include <stdlib.h>

int mins(int x, int y, int z)
{
int min = x;
int max =x;
if (y < min) min = y;
if (z < min) min = z;
if (y > max) max = y;
if (z > max) max = z;
return max-min;
}

int main(){
int H,W,i,j,k,l,S;
scanf("%d %d",&H,&W);

j=(H/3)*W;
k=(H-H/3)/2*W;
l=W*H-j-k;
S=mins(j,k,l);

for(i=0;i<2;i++){
j=(H/3+i)*W;
k=W/2*(H-H/3-i);
l=W*H-j-k;
if(mins(j,k,l)<S) S=mins(j,k,l);
}

for(i=0;i<2;i++){
j=(W/3+i)*H;
k=(W-W/3-i)/2*H;
l=W*H-j-k;
if(mins(j,k,l)<S) S=mins(j,k,l);
}
  
for(i=0;i<2;i++){
j=(W/3+i)*H;
k=H/2*(W-W/3-i);
l=W*H-j-k;
if(mins(j,k,l)<S) S=mins(j,k,l);
}

printf("%d\n",S);  
return 0;
}