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)

#define _USE_MATH_DEFINES
#include <bits/stdc++.h>
using namespace std;

#define all(x) (x).begin(), (x).end()
#define dbl(k, x) fixed << setprecision(k) << (x)

template <typename _T>
inline void _DBG(const char *s, _T x) {
  cerr << s << " = " << x << "\n";
}
template <typename _T, typename... args>
void _DBG(const char *s, _T x, args... a) {
  while (*s != ',') cerr << *s++;
  cerr << " = " << x << ',';
  _DBG(s + 1, a...);
}

#define _upgrade                \
  ios_base::sync_with_stdio(0); \
  cin.tie(0);                   \
  cout.tie(0);
#define DBG(...) _DBG(#__VA_ARGS__, __VA_ARGS__)

// ********************** CODE ********************** //

int main() {
  _upgrade;
  long long a, b, c, d;
  cin >> a >> b >> c >> d;
  long long res = a * c;
  res = max(res, a * d);
  res = max(res, b * c);
  res = max(res, b * d);
  cout << res << "\n";
  return 0;
}