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: Accepted)

#include"bits/stdc++.h"

#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define REP(i, n) for (int i = 1; i <= (int)(n); i++)
#define all(v) v.begin(), v.end()
#define rall(v) v.rbegin(), v.rend()
using namespace std;
using ll = long long;
using pi = pair<int, int>;
const ll INF = 1LL << 60;

int main() {
	ll a, b, c, d,tmp=0;
	cin >> a >> b >> c >> d;
	if (a < 0 && 0 <= b || c < 0 && 0 <= d)tmp = 0;
	else tmp = a * c;
	cout << max(max(max(max(a*c,a*d),b*c),b*d),tmp) << endl;


	return 0;
}