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)

/*#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
#pragma GCC optimization("unroll-loops")
*/

#include <bits/stdc++.h>
using namespace std;
#define pi acos(-1)
#define IOS ios_base::sync_with_stdio(0); cin.tie(); cout.tie();
#define fi first
#define se second
#define pf push_front
#define pb push_back
#define eb emplace_back
#define mp make_pair
#define debug(x) cerr<<#x<<": "<<(x)<<'\n'
typedef long double ld;
typedef long long ll;
typedef unsigned long long ull;
const ll mod = 1e9+7;
const ll inf = 0x3f3f3f3f;
const ll nax = 0;	

int main(){		
	IOS
	ll a, b, c, d; cin >> a >> b >> c >> d;
	if((a<=0 && b<=0) || (c<=0 && d<=0)){
		if(a<0 && b<0){
			if(c<0){
				cout << c*a << '\n';
			}
			else{
				cout << b*c << '\n';
			}
		}
		else{
			if(a<0){
				cout << c*a << '\n';
			}
			else{
				cout << d*a << '\n';
			}
		}	
	}
	else{
		cout << b*d << '\n';
	}
}