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: Wrong Answer)

#include <stdio.h>

long long min(long long a, long long b);
long long max(long long a, long long b);
long long ab(long long a);

int main(void){
    int H, W;
    scanf("%d %d", &H, &W);
    int sh, lo;

    if((unsigned long long)H*(unsigned long long)W%3 == 0){
        printf("0\n");
        return 0;
    }

    if(H>W){
        sh = W;
        lo = H;
    }
    else{
        sh = H;
        lo = W;
    }

    int a, b=sh/2;
    if(sh%2 == 0){
        a = b;
    }
    else{
        a = b + 1;
    }
    
    int i=lo/2;
    long long d=H*W;

    while(i<lo){
        long long S1=a*i, S2=b*i, S3=sh*(lo-i);
        if(a==b){
            d=min(d, ab(S3-S1));
            //printf("d = %d\n", d);
        }
        else{
            long long la = max(ab(S3-S1), ab(S3-S2));
            d=min(d, la);
        }
        i++;
    }
        
    printf("%lld", d);
    return 0;
}

long long min(long long a, long long b){
    if(b>a){
        return a;
    }
    else{
        return b;
    }
}

long long max(long long a, long long b){
    if(b>a){
        return b;
    }
    else{
        return a;
    }
}

long long ab(long long a){
    if(a>=0){
        return a;
    }
    else{
        return a *(-1);
    }
}