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>
typedef long long lint;
using namespace std;

lint a, b, c, d;

template <class T>
bool chmax(T &a, const T &b) {
    if (a < b) {
        a = b;
        return 1;
    }
    return 0;
}

bool contain_x() {
    return (a <= 0 && b >= 0) || (c <= 0 && d >= 0);
}

int main() {
    cin >> a >> b >> c >> d;
    lint ans = LLONG_MIN;
    chmax(ans, a * c);
    chmax(ans, a * d);
    chmax(ans, b * c);
    chmax(ans, b * d);
    if (contain_x()) {
        chmax(ans, 0LL);
    }
    cout << ans << endl;
}