Problem:
Score : 200 points

Problem Statement

Given are integers a,b,c and d.
If x and y are integers and a \leq x \leq b and c\leq y \leq d hold, what is the maximum possible value of x \times y?

Constraints

-10^9 \leq a \leq b \leq 10^9

-10^9 \leq c \leq d \leq 10^9

All values in input are integers.

Input

Input is given from Standard Input in the following format:

a b c d

Output

Print the answer.

Sample Input 1

1 2 1 1

Sample Output 1

2

If x = 1 and y = 1 then x \times y = 1.
If x = 2 and y = 1 then x \times y = 2.
Therefore, the answer is 2.

Sample Input 2

3 5 -4 -2

Sample Output 2

-6

The answer can be negative.

Sample Input 3

-1000000000 0 -1000000000 0

Sample Output 3

1000000000000000000

Solution in C (GCC 9.2.1) (judge verdict: Accepted)

#include<stdio.h>
int main(void){
    long a,b,c,d;
    scanf("%ld%ld%ld%ld",&a,&b,&c,&d);
    if (a >= 0)
    {
        if (c >= 0)
        {
            printf("%ld\n",b *d);
        }else if (c < 0)
        {
            if (d >= 0)
            {
                printf("%ld\n",b * d);
            }else if (d < 0)
            {
                printf("%ld\n",a * d);
                
            }
            
            
        }
        
        
    }

    else if (a < 0)
    {
        if (b >= 0)
        {
            if (c >= 0)
            {
                printf("%ld\n",b * d);
            }
            else if (c < 0)
            {
                if (d >= 0)
                {
                    if ((a * c) >= (b * d))
                    {
                        printf("%ld\n",a * c);
                        
                    }
                    else if ((a * c) < (b * d))
                    {
                        printf("%ld\n",b * d);

                    }
                    
                    
                }
                else if (d < 0)
                {
                    printf("%ld\n",a * c);

                }
                
                
            }
            
            
        }
        else if (b < 0)
        {
            if (c >= 0)
            {
                printf("%ld\n",b * c);
            }
            else if (c < 0)
            {

             printf("%ld\n",a * c);
     
            }
            
            
        }
        
        
    }
    
    
    return 0;
}