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 Ruby (2.7.1) (judge verdict: Accepted)

A, B, C, D = gets.split.map(&:to_i)
MAX_INT = 10**9
x_list = [A, B]
x_list << A+1 if A <  MAX_INT && A < B
x_list << B-1 if B > -MAX_INT && A < B
y_list = [C, D]
y_list << C+1 if C <  MAX_INT && C < D
y_list << D-1 if D > -MAX_INT && C < D
max_val = -MAX_INT**2
x_list.each do |x|
    y_list.each do |y|
        max_val = [x*y, max_val].max
    end
end
puts max_val