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)


/*
	*	Created By: 'Present_Sir'
	*	Created On: Sunday 13 September 2020 05:37:57 PM IST
*/

#include<bits/stdc++.h>
#define IOS ios_base::sync_with_stdio(false); cin.tie(NULL)
#define all(v) v.begin(),v.end()
#define allr(v) v.rbegin(),v.rend()
#define sz(v) ((int)v.size())
#define lgt(s) ((int)s.length())
#define int long long
using namespace std;

class Solver{
    public:
        void test(){
            //int t; cin>>t; for(int i=1; i<=t; ++i)
            {
                //cout<<"Case #"<<i<<": ";
                this->solve();
            }
        }
        void solve(){
            int a,b;
            cin>>a>>b;

            int c,d;
            cin>>c>>d;

            int mx = a*c;
            mx = max(mx, a*d);
            mx = max(mx, b*c);
            mx = max(mx, b*d);
            if( (a<=0 and b>=0) or ( c<=0 and d>=0) ){
                mx = max(mx, 0LL);
            }
            cout<<mx<<endl;
        }
};


int32_t main(){
    IOS;
    Solver S;
    S.test();
    return 0;
}