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: Wrong Answer)

// #define _GLIBCXX_DEBUG
#include <bits/stdc++.h>
using namespace std;
using namespace std::chrono;

#define ll long long
#define ull unsigned long long
#define lld long double
#define FIO ios_base::sync_with_stdio(false);cin.tie(NULL);
#define pb push_back
#define eb emplace_back
#define ff first
#define ss second
#define vt vector
#define vll vt<ll>
#define pll pair<ll,ll>
#define vpll vt<pll>
#define vvll vt<vll>
#define all(v) v.begin(),v.end()
#define FOR(i,n) for(ll i=0;i<n;i++)
#define ffo(i,a,b) for(ll i=a;i<=b;i++)
#define rfo(i,a,b) for(ll i=a;i>=b;i--)
#define space cout<<"\n\n";
#define endl '\n'
template <typename T> using mxpq = priority_queue<T>;
template <typename T> using mnpq = priority_queue<T, vt<T>, greater<T>>;
#define fps(x,y) fixed<<setprecision(y)<<x
#define merg(a,b,c) set_union(a.begin(),a.end(),b.begin(),b.end(),inserter(c,c.begin()))
#define set_ar(arr,v) memset(arr,v,sizeof(arr))
#define go_t int testcases; cin>>testcases; ffo(caseno,1,testcases)

#define ctime auto start = high_resolution_clock::now()
#define etime auto stop = high_resolution_clock::now()
#define ptime auto z1z = duration_cast<microseconds>(stop-start); cout<<"Time elapsed : "<<z1z.count()<<" microseconds\n"

const ll mod = 1e9 + 7;
const ll N = 6e4 + 6;
const ll maxN = 1e5 + 5;
const ll MAX_SIZE = 2e6 + 6;
const ll INF = 0x3f3f3f3f3f3f3f3fll;
const double PI = 3.14159265359;

int dx[4] = { -1, 0, 1, 0};
int dy[4] = {0, 1, 0, -1};
// up, right, down, left
//int dx [] = {+1,-1,+0,+0,-1,-1,+1,+1};///Eight Directions
//int dy [] = {+0,+0,+1,-1,+1,-1,-1,+1};///Eight Directions
//int dx[]={-2,-2,-1,1,-1,1,2,2};///Knight moves
//int dy[]={1,-1,-2,-2,2,2,-1,1};///Knight moves

ll powerM(ll x, ll y, ll M = mod) { // default argument
	ll v = 1; x = x % M; while (y > 0) {if (y & 1)v = (v * x) % M; y = y >> 1; x = (x * x) % M;} return v;
}

ll power(ll x, ll y) {
	ll v = 1; while (y > 0) {if (y & 1)v = v * x; y = y >> 1; x = x * x;} return v;
}

int main()
{
#ifndef ONLINE_JUDGE
	freopen("in4.txt", "r", stdin);
	freopen("out1.txt", "w", stdout);
#endif
	FIO
	ll a,b,c,d;
	cin>>a>>b>>c>>d;
	cout<<b*d;
	return 0;
}