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;

#define pb push_back
#define all(a) begin(a), end(a)
#define deb(x) cout << #x << " = " << x << endl
#define debu(x, y) cout << "( " << #x << ", " << #y << " ) = ( " << x << ", " << y << " )" << endl
#define int long long

typedef long long ll;
ll mod=1000000007;

template <typename T> vector<T> readVector(int n) { vector<T> res(n); for (int i = 0 ; i < n ; i++) cin >> res[i]; return res; }

signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    int a, b, c, d;
    cin >> a >> b >> c >> d;
    int ans;

    ans = b*d;
    //if( d < 0) {
        ans = max(ans, d*a);
    //}
    ans = max(ans, c*a);
    ans = max(ans, (min(c+1, d)*a) );
    ans = max(ans, (min(a+1, b)*c) );
    ans = max(ans, (min(c+1, d)*min(a+1, b)) );

    cout << ans << endl;

    return 0;
}