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;
#define FOR(i, a, b) for(int i = (a); i < (b); ++i)
#define REP(i, n) for(int i = 0; i < (n); ++i)
#define REPr(i, n) for(int i = (n)-1; i >= 0; --i)
#define FORq(i, m, n) for(int i = (m); i <= (n); ++i)
#define FORqr(i, m, n) for(int i = (n); i >= (m); --i)
#define MP make_pair
#define SIN(x, S) (S.count(x) != 0)
#define M0(x) memset(x, 0, sizeof(x))
#define FILL(x, y) memset(x, y, sizeof(x))
#define MM(x) memset(x, -1, sizeof(x))
#define ALL(x) (x).begin(), (x).end()
#define DB(x) cerr << #x << " = " << x << endl
#define DB2(x, y)                                                              \
    cerr << "(" << #x << ", " << #y << ") = (" << x << ", " << y << ")\n";
#define DEBUG                                                                  \
    int x12345;                                                                \
    cin >> x12345;
typedef pair<int, int> PII;
typedef pair<long long, long long> PLL;
typedef vector<int> VI;
typedef vector<VI> VVI;
typedef vector<long long> VL;
typedef long long ll;
typedef long long integer;
const long long MOD = 1e9 + 7;

///////////////////////////////////////////////
// for template
template <class T> inline bool chmin(T &a, T b) {
    if(a > b) {
        a = b;
        return true;
    }
    return false;
}
template <class T> inline bool chmax(T &a, T b) {
    if(a < b) {
        a = b;
        return true;
    }
    return false;
}
///////////////////////////////////////////////
/// ⸜( ⌓̈ )⸝ ///
int main() {
    ll a,b,c,d;
    cin >> a >> b >> c >> d;
    ll ans = max({a*c,a*d,b*c,b*d});
    cout << ans << endl;
}