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 <stdio.h>
#include <math.h>
#include <stdlib.h>

int main(){
    int a,b,c,d;
    int output;
    scanf("%d %d %d %d", &a, &b, &c, &d);
    

    if(abs(a)==abs(b) && abs(c)>abs(d)){
        output = abs(a)*abs(c);
    }
    else if(abs(a)==abs(b) && abs(c)<abs(d)){
        output = abs(a)*abs(d);
    }
    else if(abs(a)>abs(b) && abs(c)==abs(d)){
        output = abs(a)*abs(c);
    }
    else if(abs(a)<abs(b) && abs(c)==abs(d)){
        output = abs(b)*abs(c);
    }
    else if(abs(a)==abs(b) && abs(c)==abs(d)){
        output = abs(a)*abs(c);
    }
    else if(abs(a)>abs(b) && abs(c)>abs(d)){
        output = abs(a)*abs(c);
    }
    else if(abs(a)<abs(b) && abs(c)<abs(d)){
        output = abs(b)*abs(d);
    }
    else if(abs(a)>abs(b) && abs(c)<abs(d)) {
        output = abs(a)*abs(d);
    }
    else if(abs(a)<abs(b) && abs(c)>abs(d)) {
        output = abs(b)*abs(c);
    }
    
    //printf("output:%1.1f\n", output);
    printf("%d\n", output);

    return 0;
}