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>
#define mod 1000000007
using namespace std;

int main(){
	cin.tie(0),ios::sync_with_stdio(false);
	int n,k; cin>>n>>k;
	vector<int>a,ma;
	for(int i=0;i<n;++i){
		int A; cin>>A;
		if(A>=0)a.emplace_back(A);
		else ma.emplace_back(A);
	}
	bool yes=false;
	for(int i=0,lim=k/2*2;i<=lim;i+=2){
		if(ma.size()>=i&&a.size()>=k-i){
			yes=true;
			break;
		}
	}
	int64_t ans=1;
	if(yes){
		priority_queue<int>q;
		for(auto&&i:a)q.push(i);
		for(auto&&i:ma)q.push(abs(i));
		for(int i=0;i<k;++i){
			ans*=q.top(); q.pop();
			ans%=mod;
			if(ans<0)ans+=mod;
		}
	}
	else{cout<<"とおった\n"s;
		unordered_map<int,int>m;
		for(auto&&i:a)++m[i];
		priority_queue<int,vector<int>,greater<int>>mq;
		for(auto&&i:a)mq.push(i);
		for(auto&&i:ma)mq.push(abs(i));
		for(int i=0;i<k;++i){
			int t=mq.top(); mq.pop();
			if(m[t]>0)--m[t],ans*=t;
			else ans*=-t;
			ans%=mod;
			if(ans<0)ans+=mod;
		}
	}
	cout<<ans<<"\n"s;
}