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>
using namespace std;


void solve(long long a, long long b, long long c, long long d){
    long long max = -1000000000000000000;



    // for(long long i = a; i <= b; i++) {
    //     for(long long j = c; j <= d; j++) {
    //         long long buf = i * j;
    //         if(buf > max) {
    //             max = buf;
    //         }
    //     }
    // }
    if(a * d > max) {
        max = a * d;
    }
    if(a * c > max) {
        max = a * c;
    }
    if(b * c > max) {
        max = b * c;
    }
    if(b * d > max) {
        max = b * d;
    }



    cout << max << endl;
}

// Generated by 1.1.7.1 https://github.com/kyuridenamida/atcoder-tools  (tips: You use the default template now. You can remove this line by using your custom template)
int main(){
    long long a;
    scanf("%lld",&a);
    long long b;
    scanf("%lld",&b);
    long long c;
    scanf("%lld",&c);
    long long d;
    scanf("%lld",&d);
    solve(a, b, c, d);
    return 0;
}