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<string.h>
#include<math.h>
#include<limits.h>
#include<stdbool.h>
#define e 1000000007
#define PI 3.141592653589793

typedef  long long ll;

int int_cmp(const void *a, const void *b){ 
    const int *ia = (const int *)a;
    const int *ib = (const int *)b;
    return *ia  - *ib; 
} 


ll int_cmp1(const void *a, const void *b){ 
    const int *ia = (const int *)a;
    const int *ib = (const int *)b;
    return *ia  - *ib; 
} 



ll max(ll a,ll b){
	if(a<b)
		a=b;
	return a;	
}

ll min(ll a,ll b){
	if(a>b)
		a=b;
	return a;	
}

int absolu(int n){
	if(n<0)
		n*=-1;
	
	return n;	
}


void solve(){ 
	ll a,b,c,d,ans;
	scanf("%lld%lld%lld%lld",&a,&b,&c,&d);
	ll 	x=max(a*c,a*d);
	ll y=max(b*c,b*d);
	ans = max(x,y);
	printf("%lld\n",ans);
	
}

int main(){
 /* int t;
	scanf("%d",&t);
	
	while(t--)*/
		solve();
	
	return 0;
}