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 "bits/stdc++.h"
#define rep(i,n) for(int i = 0; i < (n); i++)
using ll = long long;
using namespace std;
#define SIZE_OF_ARRAY(array) (sizeof(array)/sizeof(array[0]))
const int MX = 1000000000;

int main() {
	ll a, b, c, d;
	cin >> a >> b >> c >> d;
	ll ans = 0;
	
	if (a >= 0 && c >= 0) {
		ans = b * d;
	}
	else if (b <= 0 && d <= 0) {
		ans = a * c;
	}
	else if (a >= 0 && d <= 0) {
		ans = a * d;
	}
	else if (b <= 0 && c <= 0) {
		ans = b * d;
	}
	else if (a <= 0 && b >= 0 && c <= 0 && d <= 00) {
		int x1 = a * c;
		int x2 = b * d;
		if (x1 > x2) {
			ans = x1;
		}
		else {
			ans = x2;
		}
	}
	cout << ans << endl;
}