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: Time Limit Exceeded)

                           //*****Jai Mata di*****//

#include<bits/stdc++.h>
using namespace std;
#define Nitroboost  ios_base::sync_with_stdio(0);cin.tie(NULL);cout.tie(NULL)
#define ll long long int 
const int mod=1e9+7;
const int N=2e5+15;
#define debug cout<<"Good uptill here"<<endl<<flush;
#define deb(x) cout<<#x<<":"<<x<<endl<<flush;


vector<ll>pos,neg;

bool comp(ll a, ll b)
{
    return a>b;
}

int main()
{
    Nitroboost;
    int n,k;
    cin>>n>>k;
    
    ll ans=1;
    for(int i=0;i<n;i++)
    {
        ll t;
        cin>>t;
        if(t>0)
            pos.push_back(t);
        else
            neg.push_back(t);
            
  
    }
 
       if(pos.size()!=0)
       {
        
         if(k%2!=0)
            {
                sort(pos.begin(),pos.end());
                ans=((ans*pos.back())%mod+mod)%mod;
                pos.pop_back();
                k-=1;
            }
            
        vector<ll>mul;
         sort(pos.begin(),pos.end(),comp);
         sort(neg.begin(),neg.end());
         
        for(int i=0;(i+1)<pos.size();i=i+2)
            mul.push_back(pos[i]*pos[i+1]);
        
        for(int i=0;(i+1)<neg.size();i=i+2)
            mul.push_back(neg[i]*neg[i+1]);
        
        if(pos.size()%2!=0 && neg.size()%2!=0)
            mul.push_back(pos.back()*neg.back());
        
        sort(mul.begin(),mul.end(),comp);
        for(int i=0;i<k/2;k++)
            ans=(ans*mul[i])%mod;
        cout<<ans;
    }
     else
       {
            sort(neg.begin(),neg.end());
            if(k%2!=0)
            {
                reverse(neg.begin(),neg.end());
            }
            for(int i=0;i<k;i++)
            {
                ans=(ans*neg[i]+mod)%mod;
            }
            if(ans<0)
                ans=(ans+mod)%mod;
            cout<<ans<<endl;
       }
}