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>
using namespace std;

#define rep(i, n) for(int i = 0; i < (int)(n); i++)
using ll = long long;
int mod = 1000000000+7;

int main(){

	int n,k;
	cin>>n>>k;
	int a[n];
	rep(i,n)cin>>a[i];

	sort(a,a+n);
	vector<int> neg,pos,zero;
	rep(i,n){
		if(a[i]<0)neg.push_back(a[i]);
		else if(a[i]==0)zero.push_back(a[i]);
		else pos.push_back(a[i]);
	}
	int x=pos.size(),y=neg.size();
	if(x+y<k){
		cout<<0<<endl;
		return 0;
	}

	ll ans=1;
	int p,q;
	if((x+y==k && y%2==1 && x%2==0) || (x==0 && k%2==1)){
		p=0,q=y-1;
		for(int i=k;k>0;k--){
			if(p<x && q>=0){
				if(pos[p]<abs(neg[q])){
					ans=(ans*pos[p])&mod;
					p++;
				}
				else {
					ans=(ans*neg[q])%mod;
					q--;
				}
			}
			else if(p>=x){
				ans=(ans*neg[q])%mod;
				q--;
			}
			else if(q<0){
				ans=(ans*pos[p])%mod;
				p++;
			}
		}
	}
	else {
		p=x-1,q=0;
		for(int i=0;i<k;i++){
			 if(i==k-1){
				if(q%2==1 && q!=y-1)ans=(ans*neg[q])%mod;
				else ans=(ans*pos[p])%mod;
			}
			else if(p>0 && q<y-1){
				if(pos[p]>abs(neg[q])){
                    ans=(ans*pos[p])&mod;
                    p--;
                }
                else {
                    ans=(ans*neg[q])%mod;
                    q++;
                }
			}
			else if(p<0){
				ans=(ans*neg[q])%mod;
				q++;
			}
			else if(q>y-1){
				ans=(ans*pos[p])%mod;
				p--;
			}
		}
	}


	if(ans<0)ans+=mod;
	cout<<ans<<endl;

	return 0;
}