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 pb push_back
#define eb emplace_back
#define mk make_pair
#define fi first
#define se second
#define mset(a, b) memset(a, b, sizeof(a))
#define DBG(x) cout << "[" << #x << "]: " << x << endl
using ll = long long;
using pii = pair<int, int>;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
#ifdef _WIN32
#define getchar_unlocked _getchar_nolock
#endif

template<class Ty> Ty randint(Ty a, Ty b) { return uniform_int_distribution<Ty>(a, b)(rng); }
template<class num> inline void rd(num& x) {
	char c, neg = 0; while(isspace(c = getchar_unlocked()));
	if(!isdigit(c)) neg = (c == '-'), x = 0;
	else x = c - '0';
	while(isdigit(c = getchar_unlocked())) x = (x << 3) + (x << 1) + c - '0';
	x = neg ? -x : x; }
template <class... Args> inline void rd(Args&... args) { (rd(args), ...); }

const int MAXN = 2e5 + 5, INF = 0x3f3f3f3f;

int main() {
	ll a, b, c, d; rd(a, b, c, d);

	ll ans = LLONG_MIN;
	ans = max(ans, a * c);
	ans = max(ans, b * c);
	ans = max(ans, a * d);
	ans = max(ans, b * d);
	printf("%lld\n", ans);
}