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)

                           //*****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<int>adj[N];
int h[N];
ll ans=0;
priority_queue<int,vector<int>,greater<int> >q;
vector<pair<int,int> >L,R;
int rsum[100],dsum[100];
int dp[100];


int main()
{
    Nitroboost;
    int n,k;
    cin>>n>>k;
    ll a[n+1];
    for(int i=1;i<=n;i++)
    {
        cin>>a[i];
    }
    
    sort(a+1,a+n+1);
    
    ll l[n+1],r[n+2];
   
    l[0]=1;
    r[n+1]=1;
    
     for(int i=1;i<=n;i++)
     {
         l[i]=(a[i]*l[i-1])%mod;
     }
     
     for(int i=n;i>=1;i--)
     {
         r[i]=(r[i+1]*a[i])%mod;
     }
     
     ll ans=-1e15;
     for(int i=0;i<=k;i++)
     {
         if(n-k+i+1>=0)
         ans=max(ans,(l[i]*r[n-k+i+1]+mod)%mod);
     }
     cout<<ans;
    }