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<bits/stdc++.h>
using namespace std;
int main()
{
    long long a,b,c,d;
    cin>>a>>b>>c>>d;
    long long ans=1;
    int option2=1;
    if(a<0 && b<0)
    {
        if(c>0 && d>0)
        {
            ans=max(a,b)*min(c,d);
        }
        else
        {
            ans=min(a,b)*min(c,d);
        }
    }
    else if(a>0 && b<0)
    {
        if(c>0 && d>0)
        {
            ans=max(c,d)*a;
        }
        else if(c>=0 && d<=0)
        {
            ans=max(a*c,b*d);
        }
        else if(c<=0 && d>=0)
        {
            ans=max(a*d,c*b);
        }
        else if(c<0 && d<0)
        {
            ans=b*min(c,d);
        }
    }
    else if(a<=0 && b>=0)
    {
        if(c>0 && d>0)
        {
            ans=max(c,d)*b;
        }
        else if(c>=0 && d<=0)
        {
            ans=max(c*b,a*d);
        }
        else if(c<=0 && d>=0)
        {
            ans=max(c*a,b*d);
        }
        else if(c<0 && d<0)
        {
            ans=a*min(c,d);
        }
    }
    else if(a>0 && b>0)
    {
        if(c>0 && d>0)
        {
            ans=max(c,d)*max(a,b);
        }
       else
       {
           ans=max(c,d)*min(a,b);
       }

    }
    cout<<ans<<endl;
}