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)


// Problem : E - Multiplication 4
// Contest : AtCoder - AtCoder Beginner Contest 173
// URL : https://atcoder.jp/contests/abc173/tasks/abc173_e
// Memory Limit : 1024 MB
// Time Limit : 2000 ms
// Powered by CP Editor (https://github.com/cpeditor/cpeditor)

#include <bits/stdc++.h>
#define ll long long
#define fast ios_base::sync_with_stdio(false);cin.tie(NULL)
#define fr(i,a,b) for(ll i=a;i<b;i++)
#define frr(i,a,b) for(int i=a;i>=b;i--)
//vector
#define pb push_back
#define vi vector<int>
#define vl vector<ll>
#define pp pop_back
//pair
#define pi pair<int ,int >
#define mk make_pair
#define ff first
#define ss second
//set
#define si set<int>
#define sit set<int>::iterator 
using namespace std;
int main()
{
fast;
ll n,m;cin>>n>>m;
ll p=0,ng=0,ze=0;
vl vp,vn;
ll x,mod=1e9+7;
fr(i,0,n)
{
	cin>>x;
	if(x>0)
	{
		p++;
		vp.pb(x);
	}
	else if(x<0)
	{
		ng++;
		vn.pb(x);
	}
	else
	{
		ze++;
	}
}

sort(vn.begin(),vn.end());
sort(vp.begin(),vp.end());
ll t=ng;
if(t%2) t--;
ll tp=p;
if(tp%2) tp--;
if(p>=m || ((tp+t)>=m && m%2==0))
{
	fr(i,0,ng)
	{
		vp.pb(abs(vn[i]));
	}
	sort(vp.begin(),vp.end(),greater<ll>());
	ll ans=1;
	
	fr(i,0,m)
	{
		ans*=vp[i];
		ans%=mod;
	}
	cout<<ans;
}
else if(ze>0) cout<<0;
else
{
	ll i=0,j=ng-1,ans=1;
	while(m--)
	{
		if(i<p && j>=0)
		{
			if((abs(vn[j]))<vp[i])
			{
				ans*=abs(vn[j]);
				ans%=mod;
				j--;
			}
			else
			{
				ans*=vp[i];
				i++;
				ans%=mod;
			}
		}
		else if(i<p)
		{
			ans*=vp[i];
				i++;
				ans%=mod;
		}
		else
		{
			    ans*=abs(vn[j]);
				ans%=mod;
				j--;
		}
	}
	ans*=-1;
	ans+=mod;
	ans%=mod;
	cout<<ans;
}
return 0;
}