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>
using namespace std;
using ll = long long;
using ld = long double;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
//二項係数の計算
int nCr(int A, int B) {
    ll ans = 1;
    for (int i = 0; i < B; i++) {
        ans *= A - i;
    }
    for (int i = 0; i < B; i++) {
        ans /= i + 1;
    }
    return ans;
}
int main() {
    ll a,b,c,d;
    cin>>a>>b>>c>>d;
    ll ans=0;
    if(0<a&&0<c){
        ans=b*d;
    }
    else if(a<0&&c<0){
        ans=max(a*c,b*d);
    }
    else{
        ans=max(b*d,a*c);
    }
    cout<<ans<<endl;
}