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 (Clang 10.0.0) (judge verdict: Wrong Answer)

#include<stdio.h>
#include<math.h>
int main() {

	unsigned long long int max = -1;
	long long int a, b, c, d, x, y;

	scanf("%lld%lld%lld%lld", &a, &b, &c, &d);

	if ((a < 0 && b < 0) || (c < 0 && d < 0)) {
		if (abs(a) < abs(b))
			x = abs(a);
		else
			x = abs(b);
		if (abs(c) < abs(d))
			y = abs(c);
		else
			y = abs(d);

		max = x * y;
		printf("-%lld", max);	
	}

	else {
		if ((a >= 0 && c >= 0) || (a < 0 && c < 0)) {
			max = a * c;
		}
		if ((a >= 0 && d >= 0) || (a < 0 && d < 0)) {
			if (max < a * d)
				max = a * d;
		}
		if ((b >= 0 && c >= 0) || (b < 0 && c < 0)) {
			if (max < b * c)
				max = b * c;
		}
		if ((b >= 0 && d >= 0) || (b < 0 && d < 0)) {
			if (max < b * d)
				max = b * d;
		}
		printf("%lld", max);
	}

	return 0;
}