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<bits/stdc++.h>
using namespace std;
#define ll long long
#define Mod 1000000007
#define Max 1000001
#define vt vector<long long>

void func(){
    ll arr[4];
    for(int i=0; i<4; i++){
        cin>>arr[i];
    }
    
    ll val1=arr[0]*arr[2];
    ll val2=arr[0]*arr[3];
    ll val3=arr[1]*arr[2];
    ll val4=arr[1]*arr[3];
    
    cout<<max(val1, max(val2, max(val3, val4)))<<endl;
    
    
}

int main(){
 
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
 

// 
// int t;
// cin >> t;
// for(ll i=1; i<=t; i++)
// {
    func();
//    cout<<endl;
//cout<<func()<<endl;
 
//if(func())
//    cout<<"Yes"<< endl;
//else
//    cout <<"No"<<endl; 
//cout<<"Case"<<" "<<"#"<<i<<":"<<" "<<func()<<endl;
// }
return 0;
}