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 <stdio.h>
#include <inttypes.h>

int64_t neko;

void mofu(int64_t nyan) {
	if (neko < nyan) neko = nyan;
}

int main(void) {
	int64_t a, b, c, d;
	int i, j;
	if (scanf("%" SCNd64 "%" SCNd64 "%" SCNd64 "%" SCNd64, &a, &b, &c, &d) != 4) return 1;
	neko = a * c;
	for (i = 0; i < 5 && a + i <= b; i++) {
		for (j = 0; j < 5 && c + j <= d; j++) mofu((a + i) * (c + j));
		for (j = 0; j < 5 && d - j >= c; j++) mofu((a + i) * (d - j));
	}
	for (i = 0; i < 5 && b - i >= a; i++) {
		for (j = 0; j < 5 && c + j <= d; j++) mofu((b - i) * (c + j));
		for (j = 0; j < 5 && d - j >= c; j++) mofu((b - i) * (d - j));
	}
	printf("%" PRId64 "\n", neko);
	return 0;
}