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>
#include<iostream>
#include<algorithm>
#include <cmath> 
using namespace std;
const int p=31;
const int m=1e9+7;
typedef long long ll;
typedef int in;
#define int long long 
ll mod=1e9+7;
   
in main (){

    int n,k,ps=0,ng=0,ze=0;
    cin>>n>>k;
    vector<int>v(n);
    for(int i=0;i<n;++i){
    	cin>>v[i];
    	if(v[i]>0)ps++;
    	else if(v[i]==0)ze++;
    	else ng++;
    }
    if(ze!=0 && k>(ps+ng-(ng%2)) ){
    	cout<<0;
    	return(0);
    }
    sort(v.begin(),v.end());
    int ans=1;
    if(ng==n || k==n){
    	if(k==n && ze!=0){
    		cout<<0;
    		return(0);
    	}	
    	for(int i=n-1;i>=n-k;--i){
    		ans*=v[i];
    		ans=ans%m;
    	}
    	cout<<(ans+m)%m;
    	return(0);
    }
    while(k>0){
    	int i=0;
    	if(ng>=2 && k>1){
    		int _n=v[i]*v[i+1];
    		int _p=v[n-1]*v[n-2];
    		if(_n>=_p) {
    			_n%=m;
    			ans*=_n;
    			ans%=m;
    			i+=2;
    			k-=2;
    		}
    		else{
    			ans*=v.back();
    			ans%=m;
    			v.pop_back();
    			k--;
    		}

    	}
    	else{
    		ans*=v.back();
    		ans%=m;
    		v.pop_back();
    		k--;
    	}
    }
    cout<<(ans+m)%m;
   // cerr << "Time : " << 1000 * ((double)clock()) / (double)CLOCKS_PER_SEC << "ms\n"; //check time
    return(0);
	
}