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 <stdlib.h>
#include <string.h>

int compare(const void*a,const void *b) {
	return *(int*)a - *(int*)b;
}

int main() {
	long long int n, t, x, a, b, c, d, ans = 0, r[200010] = { 0 };
	char s[1000],T[1000];
	scanf("%lld%lld%lld%lld", &a,&b,&c,&d);
	//scanf("%s", T);
	//for (int i = 0; i + strlen(T) <= strlen(s); i++) {
		//int re = 0;
		///for (int j = 0; j < strlen(T); j++) if (s[i + j] != T[j])re++;
		//if (re < ans)ans = re;
	//}
	ans = a * c;
	n = a * d;
	t = b * c;
	x = b * d;
	if (ans < n)ans = n;
	if (ans < t)ans = t;
	if (ans < x)ans = x;
	else printf("%lld",ans);
	return 0;
}