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: Wrong Answer)

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

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

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

        }
        
        
    }
    
    
    return 0;
}