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: Accepted)

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<math.h>
#include<queue>
#include<stdlib.h>
#include<vector>
#include<stack>
#include<bitset>
#include<set>
#define maxn 400005
using namespace std;
#define int long long
typedef long long ll;
const ll mod=1e9+7;
int n,k;
int a[maxn];
int myabs(int now)
{
	return now>0 ? now : -now;
}
int cmp(int a,int b)
{
	return myabs(a)>myabs(b);
}
int cmp1(int a,int b)
{
	return myabs(a)<myabs(b);
}
vector<int> edge1,edge2;
long double sum1[maxn],sum2[maxn];
signed main()
{
	scanf("%lld%lld",&n,&k);
	for(int i=1;i<=n;i++)
	{
		scanf("%lld",&a[i]);
		if(a[i]<0)
		{
			edge2.push_back(-a[i]);
		}
		else if(a[i]>0)
		{
			edge1.push_back(a[i]);
		}
	}
	if(k==n)
	{
		ll gg=1;
		for(int i=1;i<=n;i++)
			gg=(gg*a[i])%mod;
		printf("%lld\n",(gg+mod)%mod);
		return 0;
	}
	sort(edge1.begin(),edge1.end(),cmp);
	sort(edge2.begin(),edge2.end(),cmp);
	
	int len1=edge1.size(),len2=edge2.size();
	for(int i=0;i<len1;i++)
		sum1[i+1]=sum1[i]+log2(edge1[i]);
	for(int i=0;i<len2;i++)
		sum2[i+1]=sum2[i]+log2(edge2[i]);
	
	double ans=-1;
	int pos=0,flag=1;
	for(int i=0;i<=len2;i+=2)
	{
		int temp=k-i;
		if(temp>len1 || temp<0) continue;
		flag=0;
		if(sum1[temp]+sum2[i]>ans)
		{
			ans=sum1[temp]+sum2[i];
			pos=i;
		}
	}
	
	if(flag==1)
	{
		for(int i=1;i<=n;i++)
		{
			if(a[i]==0)
			{
				printf("0\n");
				return 0;
			}
		}
		
		sort(edge1.begin(),edge1.end(),cmp1);
		sort(edge2.begin(),edge2.end(),cmp1);
		int len1=edge1.size(),len2=edge2.size();
		for(int i=0;i<len1;i++)
			sum1[i+1]=sum1[i]+log2(edge1[i]);
		for(int i=0;i<len2;i++)
			sum2[i+1]=sum2[i]+log2(edge2[i]);
		double ans=1e18;
		pos=0;
		for(int i=1;i<=len2;i+=2)
		{
			int temp=k-i;
			if(temp>len1 || temp<0) continue;
			if(sum1[temp]+sum2[i]<ans)
			{
				ans=sum1[temp]+sum2[i];
				pos=i;
			}
		}
		ll res=1;
		for(int i=0;i<pos;i++)
		{
			res=(res*edge2[i])%mod;
		}
		for(int i=0;i<k-pos;i++)
		{
			res=(res*edge1[i])%mod;
		}
		printf("%lld\n",(-res+mod)%mod);
	}
	else
	{
		ll res=1;
		for(int i=0;i<pos;i++)
		{
			res=(res*edge2[i])%mod;
		}
		for(int i=0;i<k-pos;i++)
		{
			res=(res*edge1[i])%mod;
		}
		printf("%lld\n",res);
	}
}