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 P = pair<int,int>;

int main() {
  int i, N, K, x, count=0, ans=0;
  long long a, b, c, d, max1, max2;
  
  cin >> a >> b >> c >> d;
  
  max1 = a*c;
  if(a*d>max1) max1=a*d;
  else if(b*c>max1) max1=b*c;
  else if(b*d>max1) max1=b*d;
  
  if(a*b<=0){
    if(max1<0) max1=0;
  }
    
  if(c*d<=0){
    if(max1<0) max1=0;
  }
  cout << max1 << endl;
}