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)

/*** akshaykumar99 ***/
 
#include<bits/stdc++.h> 
using namespace std;
 
const long long MOD = 1e9+7;
const long double PI = acos(-1);
const long long N = 2e9;
 
typedef long long ll;
typedef long double ld;

void test_case()
{
	ll a, b, c, d;
	cin >> a >> b >> c >> d;
	ll ans = a * c;
	ans = max(ans, a * d);
	ans = max(ans, b * d);
	ans = max(ans, b * c );
	cout << ans << '\n';
	
}
 
int32_t main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL); cout.tie(NULL); 
    ll tc=1;
    // cin>>tc;
    while(tc--)
        test_case();
    return 0;
}