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 <iostream>
#include <algorithm>
#include <vector>
#include <map>
#include <set>
//#include <numeric> //iota
//#include <sstream>
#include <cstring>
//#include <tuple>
//#include <random>
//#include <string>
//#include <unordered_set>
//#include <bitset>
//#include <unordered_map>
//#include <stack>
#include <cmath>
//#include <deque>
#include <queue>
//#include <list>
#include <array>
using namespace std;
#define endl "\n"
#define all(x) (x).begin(), (x).end()
#define vec vector
#define pb push_back
#define pp pop_back
#define fi first
#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define se second
#define un_map unordered_map
#define rep(i,a,n) for (int i = a; i < n; ++i) 
#define inc_rep(i,a,n,inc) for (int i = a; i < n; i += inc) 
#define couts(x) cout << x << "\n" 
#define watch(x) cout << #x << " : " << x << endl;
typedef long long ll;
typedef pair<int,int> PII;
typedef pair<long long, long long>  PLII;
template<typename T>
void print(vector<T>& v) {rep(i, 0, (int)v.size()) cout << v[i] << " ";cout << endl;}
int nxt() {int x;cin >> x;return x;}
ll nxtl() {ll x;cin >> x;return x;}
string nxts() {string x;cin >> x;return x;}
double nxtd() {double x; cin >> x; return x;}

void __print(int x) {cerr << x;}
void __print(long x) {cerr << x;}
void __print(long long x) {cerr << x;}
void __print(unsigned x) {cerr << x;}
void __print(unsigned long x) {cerr << x;}
void __print(unsigned long long x) {cerr << x;}
void __print(float x) {cerr << x;}
void __print(double x) {cerr << x;}
void __print(long double x) {cerr << x;}
void __print(char x) {cerr << '\'' << x << '\'';}
void __print(const char *x) {cerr << '\"' << x << '\"';}
void __print(const string &x) {cerr << '\"' << x << '\"';}
void __print(bool x) {cerr << (x ? "true" : "false");}

template<typename T, typename V>
void __print(const pair<T, V> &x) {cerr << '{'; __print(x.first); cerr << ','; __print(x.second); cerr << '}';}
template<typename T>
void __print(const T &x) {int f = 0; cerr << '{'; for (auto &i: x) cerr << (f++ ? "," : ""), __print(i); cerr << "}";}
void _print() {cerr << "]\n";}
template <typename T, typename... V>
void _print(T t, V... v) {__print(t); if (sizeof...(v)) cerr << ", "; _print(v...);}

#ifndef ONLINE_JUDGE
#define debug(x...) cerr << "[" << #x << "] = ["; _print(x)
#else
#define debug(x...)
#endif

ll power_mod(ll a, ll b, ll mo) {
	ll ans = 1;
	while (b) {
		if(b & 1) {
			ans = (ans % mo * a % mo) % mo;
		}
		a = (a % mo * a % mo) % mo;
		b /= 2;
	}
	return ans;
}
void INVERSE(ll nn, ll mo, vector<ll> fact, vector<ll> mul) {
	fact[0] = 1;
	for (ll i = 1; i <= nn; ++i) {
		fact[i] = (fact[i - 1] * i) % mo;
	}
	vector<ll> inv(nn);
	inv[0] = 1;
  inv[1] = 1;
	for (ll i = 2; i <= nn; ++i) {
		inv[i] = (mo - (mo / i) * inv[mo % i] % mo) % mo; 
	}
	mul[0] = 1;
	for (int i = 1; i <= nn; ++i) {
		mul[i] = (mul[i - 1] * inv[i]) % mo;
	}
}
//int const mod = 1e9 + 7;
//int z = 2e5 + 1;
//vector<ll> inv;
//vector<ll> fact(z + 1);

//ll choose (ll n, ll k) {
  //if (n < k || k < 0 || n < 0) return 0;
  //return (fact[n] % mod * (inv[k] % mod * inv[n - k] % mod) % mod) % mod; 
//}

int main() {
	IOS;
	ll a, b, c, d;
	cin >> a >> b >> c >> d;
	cout << max({b * d, a * c, a * d, b * c});
}