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>
using namespace std;
#define ll long long 
#define mod 1000000007
#define N 200005

ll n,a[N],k,b[N],c[N],B,C;
int cmp(ll a,ll b){return a>b;}

int main(){
	cin>>n>>k;
	for(int i=1;i<=n;i++){
		cin>>a[i];
		if(a[i]<0)c[++C]=-a[i];
		else b[++B]=a[i];
	}
	if(k==n){
		ll ans=1;
		for(int i=1;i<=n;i++)ans=ans*a[i]%mod;
		cout<<ans<<'\n';
		return 0;
	}
	
	sort(c+1,c+1+C,cmp);
	sort(b+1,b+1+B,cmp);
	
	if(C==n){
		ll ans=1;
		if(k%2==0){
			for(int i=1;i<=k;i++)ans=ans*(-c[i])%mod;
		}else {
			for(int i=n;i>=n-k+1;i--)ans=ans*(-c[i])%mod;
		}
		cout<<(ans+2*mod)%mod<<'\n';
	}
	
	ll ans=1,p1=1,p2=1;
	if(k%2){
		k--;
		ans=b[p1++];
	}
	while(k){
		ll x=0,y=0;
		if(p1+1<=B){
			x=b[p1]*b[p1+1];
		}
		if(p2+1<=C){
			y=c[p2]*c[p2+1];
		}
		if(x>=y){
			ans=ans*b[p1]%mod*b[p1+1]%mod;p1+=2;
		}else {
			ans=ans*c[p2]%mod*c[p2+1]%mod;p2+=2;
		}
		k-=2;
	}
	
	cout<<ans<<'\n';
}


/*
9 7
-1 -1 -2 -3 -4 -5 
0 4 4

*/