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 Java (OpenJDK 11.0.6) (judge verdict: Wrong Answer)

import java.util.*;

public class Main {
	public static void main(String[] args){
		Scanner sc = new Scanner(System.in);
		int a = sc.nextInt();
    int b = sc.nextInt();
    int c = sc.nextInt();
    int d = sc.nextInt();
    long max = 0;

//積が最大になる掛け算を考える
//0<x,0<yのときはx,yは最大値
//x<0,y<0のときはx,yは最小値がベスト
//0<x,y<0のときは、xが最小値,yが最大値
//x<0,0<yのときは、xが最大値,yが最小値

    if(0<=a && 0<=c){
      System.out.println("max = b*d;");
      max = (long)b*d;
    }else if(b<=0 && d<=0){
      System.out.println("max = a*c;");
      max = (long)a*c;
    }else if(0<=a && d<=0){
      System.out.println("max = a*d;");
      max = (long)a*d;
    }else if(b<=0 && 0<=c){
      System.out.println("max = b*c;");
      max = (long)b*c;
    }

    System.out.println(max);

	}
}