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<bits/stdc++.h>
#define ll long long
#define ld long double
#define fi first
#define se second
#define vll vector<ll>
#define pii pair<ll,ll>
#define vvll vector< vector<ll> >
#define pb push_back
#define sz(v) (ll)(v).size()
#define inf 1e18
#define md 1000000007
#define all(v) (v).begin(),(v).end()
#define rep(i,a,b) for(ll i=a;i<b;++i)
#define tel(a){cout<<a<<"\n";}
#define tell(a,b){cout<<a<<" | "<<b<<"\n";}
#define telll(a,b,c){cout<<a<<" | "<<b<<" | "<<c<<"\n";}
#define teln(v,n){cout<<"v- ";rep(o,0,n)cout<<v[o]<<" ";cout<<"\n";}
#define IOS ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
using namespace std;

#define M 100010


int main(){
    IOS;
     
    ll n,k;cin>>n>>k;
    ll a[n];
    rep(i,0,n)cin>>a[i];
    sort(a,a+n);
    ll ans=1;
    if(a[n-1]==0 and (k&1)){
    	cout<<0;return 0;
    }
    if(a[n-1]<=0 and (k&1)){
    	for(ll i=n-1;i>=n-k;--i){a[i]%=md;(ans*=a[i])%=md,(ans+=md)%=md;}
    	cout<<ans;
    	return 0;
    }
    ll i=0,j=n-1;
    if(k&1){a[j]%=md,(ans*=a[j])%=md,(ans+=md)%=md;j--,k--;}k/=2;
	rep(p,0,k){
		ll l=a[i]*a[i+1],r=a[j]*a[j-1];
		if(l>r){l%=md;(ans*=l)%=md,(ans+=md)%=md;i+=2;}
		else {r%=md;(ans*=r)%=md,(ans+=md)%=md;j-=2;}
		// tell(i,j);
	}     
    cout<<ans;
    
    return 0;
}