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;

typedef vector<int> vi;
typedef pair<int,int> pii;
#define For(i,n) for(int i = 0; i < (int) n; ++i)
#define Rep(i,n) for(int i = 1; i <= (int) n; ++i)
#define ll long long
#define lld long double
#define F first
#define S second
#define pb push_back
#define endl "\n"
#define ar array
#define INF 1e18;
#define all(v) (v).begin(),(v).end()
const int N=2e5+5,mod=1e9+7;

void solve(){
    ll a,b,c,d;
    cin>>a>>b>>c>>d;

    ll val=max({a*c,b*d,a*d,b*c});
    cout<<val<<endl;

}


int main(){
    ios_base::sync_with_stdio(0);
    cin.tie(0);cout.tie(0);
    int t=1;
    //cin>>t;
    while(t--){
        //cout<<"Case #"<<a<<": ";
        solve();
        //a+=1;
    }
}