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(){
    long long int a,b,c,d;
    scanf("%lld",&a);
    scanf("%lld",&b);
    scanf("%lld",&c);
    scanf("%lld",&d);

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