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 all(v) v.begin(), v.end()
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define intll int long long
const int INF = 1e9;
const int MOD = 1e9 + 7;

int main() {
  intll a, b, c, d ;
  cin >> a >> b >> c >> d ; 
  
  intll ans = a*c ;
  ans = max(ans,a*d) ;
  ans = max(ans,b*c) ;
  ans = max(ans,b*d) ;
  
  intll x = 0 ;
  if((a<=0&&b>=0) || (c<=0&&d>=0)) ans = max(ans,x) ; 
  cout << ans << endl ;
  

}