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: Runtime Error)

#include<bits/stdc++.h>
#define pb push_back
using namespace std;
typedef long long ll;
const ll m=1000000007,inf=1e18+5;
ll n,k,a;
vector<ll> neg,pos,nol,jnn,jp;
ll hitung()
{
	ll now=1;
	if(!jp.empty()) 
	{
		for(ll i=0;i<jp.size();i++) now=(now*jp[i])%m;
	}
	if(!jnn.empty())
	{
		for(ll i=0;i<jnn.size();i++) now=(now*jnn[i])%m;
	}
	return now%m;
}
int main()
{
	cin>>n>>k;
	for(ll i=1;i<=n;i++) 
	{
		cin>>a;
		if(a>0) pos.pb(a);
		else if(a==0) nol.pb(a);
		else neg.pb(abs(a));
	}
	sort(pos.begin(),pos.end(),greater<ll>());
	sort(neg.begin(),neg.end(),greater<ll>());
	if(pos.size()+neg.size()<k) 
	{
		cout<<"0"<<endl;
		return 0;
	}
	if(pos.size()==0&&nol.size()==0)
	{
		if(k%2==1) 
		{
			//cout<<"asu"<<endl;
			for(ll i=neg.size()-1;i>=neg.size()-k;i--) jnn.pb(neg[i]);
			ll p=-1*hitung();
			cout<<p+m<<endl;
			return 0;
		}
	}
	ll cur=k,negg=0,mp,mn;
	while(cur--)
	{
		if(pos.size()>0)
		{
			if(neg.size()>0)
			{
				if(pos[0]>=neg[0]) jp.pb(pos[0]),mp=pos[0],pos.erase(pos.begin());
				else jnn.pb(neg[0]),mn=neg[0],neg.erase(neg.begin()),negg++;
			}
			else jp.pb(pos[0]),mp=pos[0],pos.erase(pos.begin());
		}
		else jnn.pb(neg[0]),mn=neg[0],neg.erase(neg.begin()),negg++;
	}
	if(negg%2==1)
	{
		ll cek=0,now=inf;
		if(pos.size()>0)
		{
			cek=1;
			if(jp.size()>0) now=jp.back()*pos[0];
			else now=pos[0];
		}
		else 
		{
			cout<<"0"<<endl;
			return 0;
		}
		if(neg.size()>0)
		{
			if(jnn.back()*neg[0]<now) cek=-1,now=jnn.back()*neg[0];
		}
		if(cek==0) 
		{
			if(nol.size()>0) 
			{
				cout<<"0"<<endl;
				return 0;
			}
			else 
			{
				ll tmp=-1*hitung();
				cout<<tmp+m<<endl;
				return 0;
			}
		}
		else if(cek==1)
		{
			jnn.pop_back(),jp.pb(pos[0]);
		}
		else if(cek==-1)
		{
			jp.pop_back(),jnn.pb(neg[0]);
		}
		ll tmp=hitung();
		cout<<tmp%m<<endl;
	}
	else 
	{
		ll tmp=hitung();
		cout<<tmp%m<<endl;
	}
}