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>
#define rep(i,n) for (int i = 0; i < (n); ++i)
using namespace std;
using ll = long long;
using vi=vector<int>;
using P = pair<int,int>;
using Graph = vector<vector<int>>;

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    
    ll a,b,c,d;
    cin >> a >> b >> c >> d;
    if(b>0 && d>0){
        cout << b*d << "\n";
        return 0;
    }else if(b<=0 && d<=0){
        cout << a*c << "\n";
        return 0;
    }else if(a>=0 && d<0){
        cout << a*d << "\n";
        return 0;
    }else if(b<0 && c>=0){
        cout << b*c << "\n";
        return 0;
    }
    return 0;
}