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>
typedef long long ll;
using namespace std;
ll MOD = 1000000007;

#define vec              vector<int>
#define vecll            vector<ll>
#define vecd             vector<double>
#define vecst            vector<string>
#define vecb             vector<bool>
#define v2(v,n,m,init)   vector<vector<int>>  v(n, vector<int>(m, init))
#define vb2(v,n,m,init)  vector<vector<bool>> v(n, vector<bool>(m, init))
#define vll2(v,n,m,init) vector<vector<ll>>   v(n, vector<ll>(m, init))

#define rep(i,n)         for(ll i=(ll)0; i<(ll)n; i++)
#define REP(i,m,n)       for(ll i=(ll)m; i<(ll)n; i++)
#define arr(var, n)      vec var(n); rep(i,n){cin >> var[i];}
#define arrll(var, n)    vecll var(n); rep(i,n){cin >> var[i];}
#define arrst(var, n)    vecst var(n); rep(i,n){cin >> var[i];}
#define all(var)         (var).begin(), (var).end()
#define sortall(var)     sort(all(var))
#define uniqueall(v)     v.erase(unique(v.begin(), v.end()), v.end());
#define pb(var)          push_back(var)
#define prt(var)         cout << (var) << "\n"
#define prtd(n, var)     cout << fixed << setprecision(n) << (var) << "\n"
#define prtfill(n, var)  cout << setw(n) << setfill('0') << (var);
#define prt2(v1, v2)     cout << (v1) << " " << (v2) << "\n"
#define prt3(v1, v2, v3) cout << (v1) << " " << (v2) << " " << (v3) << "\n"
#define prtall(v)        rep(i,v.size()){cout<<v[i]<<(i!=v.size()-1?" ":"\n");}
template <typename T>
bool chmax(T &a, const T& b){if(a<b){a=b; return true;} return false;}
template <typename T>
bool chmin(T &a, const T& b){if(a>b){a=b; return true;} return false;}
//------------------------------------------------------------------

int main(void){
  ll a, b, c, d;
  cin >> a >> b >> c >> d;

  ll ans=-2e18;
  chmax(ans, a*c);
  chmax(ans, a*d);
  chmax(ans, b*c);
  chmax(ans, b*d);
  prt(ans);
}