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)


// Problem : B - Product Max
// Contest : AtCoder - AtCoder Beginner Contest 178
// URL : https://atcoder.jp/contests/abc178/tasks/abc178_b
// Memory Limit : 1024 MB
// Time Limit : 2000 ms
// Powered by CP Editor (https://github.com/cpeditor/cpeditor)

//#pragma GCC optimize "trapv"//  WA to RE int!!
#include <bits/stdc++.h>
#define FOR(i,n) for(int i=0;i<n;i++)
#define FORR(i,n) for(int i=n;i>=0;i--)
#define pb push_back
#define vint vector<int>
#define sint set<int>
#define mint map<int,int>
#define pint pair< int,int >
#define int long long int 
#define test int tt1234; cin>>tt1234;while(tt1234--)
#define endl "\n"
#define SI(a,n) int a[n];FOR(i,n) cin>>a[i];
#define input(a,n) int n;cin>>n; SI(a,n)
#define all(zz) zz.begin(),zz.end()
#define rep(i, begin, end) for (__typeof(end) i = (begin) - ((begin) > (end)); i != (end) - ((begin) > (end)); i += 1 - 2 * ((begin) > (end)))
#define error(x) cerr << #x <<" is " << x << endl;
#define Error(a,b) cerr<<"( "<<#a<<" , "<<#b<<" ) = ( "<<(a)<<" , "<<(b)<<" )\n";
#define precision(x) cout<<fixed<<setprecision(x);
#define fast ios_base::sync_with_stdio(false);cin.tie(NULL);
#define present(container, element) (container.find(element)!=container.end())
#define fint fast;test
#define set(a,x) memset(a,x,sizeof(a))
#define L(x)	((x)<<1)
#define R(x)	(((x)<<1)+1)
#define ft first
#define se second
#define MOD 1000000007
using namespace std;
template<typename... T> void rd(T&... args) { ((cin >> args), ...); }
template<typename... T> void pp(T... args) { ((cout << args << " "), ...); cout<<"\n"; }

void solve(){
	int a,b,c,d;
	rd(a,b,c,d);
	int ans=b*d;
	ans=max({ans,b*c,a*c,a*d})	;
	pp(ans);
	return;	
}


int32_t main() {
    // fint{
    	fast;
    	solve();
    // }
    return 0;
}