Problem:
Score : 500 points

Problem Statement

Given are N integers A_1,\ldots,A_N.

We will choose exactly K of these elements. Find the maximum possible product of the chosen elements.

Then, print the maximum product modulo (10^9+7), using an integer between 0 and 10^9+6 (inclusive).

Constraints

1 \leq K \leq N \leq 2\times 10^5

|A_i| \leq 10^9

Input

Input is given from Standard Input in the following format:

N K
A_1 \ldots A_N

Output

Print the maximum product modulo (10^9+7), using an integer between 0 and 10^9+6 (inclusive).

Sample Input 1

4 2
1 2 -3 -4

Sample Output 1

12

The possible products of the two chosen elements are 2, -3, -4, -6, -8, and 12, so the maximum product is 12.

Sample Input 2

4 3
-1 -2 -3 -4

Sample Output 2

1000000001

The possible products of the three chosen elements are -24, -12, -8, and -6, so the maximum product is -6.

We print this value modulo (10^9+7), that is, 1000000001.

Sample Input 3

2 1
-1 1000000000

Sample Output 3

1000000000

The possible products of the one chosen element are -1 and 1000000000, so the maximum product is 1000000000.

Sample Input 4

10 10
1000000000 100000000 10000000 1000000 100000 10000 1000 100 10 1

Sample Output 4

999983200

Be sure to print the product modulo (10^9+7).

Solution in C++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>

#define AndIamMr_Awesome ios::sync_with_stdio(false); cin.tie(0);cout.tie(0);

#define m 1000000007
using namespace std;

typedef long long int lli;


int main()
{
	AndIamMr_Awesome;
	lli i=1,j,t,d,n,k,x=9999999999,a,b,c,y=0,z,co,sum=1,sum1=1,p=0;
	cin>>n>>k;
	lli ar[n];
	for(i=0;i<n;i++)
	{
		cin>>ar[i];
	}
	sort(ar,ar+n);
	for(i=0;i<k;i++)
	{
	//	cout<<ar[i]%m<<"\n";
		sum=((sum%m)*(ar[i]%m))%m;
	}
	for(i=n-1;i>n-1-k;i--)
	{
		sum1=((sum1%m)*(ar[i]%m))%m;
	}
	// cout<<"sum"<<sum<<"sum1"<<sum1<<" ";
	if(sum1>sum)
	{
		if(sum1<0)
		{
			cout<<m+(sum1%m);
		}
		else{
			cout<<sum1%m;
		}
	}
	else if(sum1<sum){
		if(sum<0)
		{
			cout<<m+(sum%m);
		}
		else{
			cout<<sum%m;
		}
	}
	else{
		if(sum<0)
		{
			cout<<m+(sum%m);
		}
		else{
			cout<<sum%m;
		}
	}
	
	
	return 0;
}