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)

//#define _GLIBCXX_DEBUG
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
#define pp pair<int,int>
#define rep(i,n) for(int (i)=0;(i)<(n);(i)++)
#define ld long double
#define al(a) (a).begin(),(a).end()
#define mk make_pair
#define check cout<<"?"<<endl;

ll MOD=1000000007;
ll mod=998244353;
int inf=1000001000;
ll INF=1e18+5;

int main(){
    ll n,k,pcnt=0,ncnt=0; cin>>n>>k;
    vector<ll> p,ne;
    rep(i,n){
        ll a; cin>>a;
        if(a>0){
            pcnt++; p.push_back(a);
        }
        else{
            ncnt++; ne.push_back(a);
        }
    }
    if(pcnt+ncnt<k){
        cout<<0<<endl;
        return 0;
    }
    if(pcnt==0){
        sort(al(ne)); 
        if(k%2) reverse(al(ne));
        ll ans=1;
        rep(i,k) ans=ans*(ne[i]+MOD)%MOD;
        cout<<(ans+MOD)%MOD<<endl;
        return 0;
    }
    if(ncnt==0){
        sort(al(p)); reverse(al(p));
        ll ans=1;
        rep(i,k) ans=ans*p[i]%MOD;
        cout<<(ans+MOD)%MOD<<endl;
        return 0;
    }
    ll pr=min(pcnt,k),nr=max(0LL,k-pcnt);
    if(nr%2){
        if(nr==ncnt){
            ll ans=1;
            rep(i,pr) ans=ans*p[i]%MOD;
            rep(i,nr) ans=ans*(ne[i]+MOD)%MOD;
            cout<<(ans+MOD)%MOD<<endl;
            return 0;
        }
        else{
            pr--; nr++;
        }
    }
    sort(al(ne));
    sort(al(p)); reverse(al(p));
    while(1){
        if(pr<2 || nr+2>ncnt) break;
        if(ne[nr]*ne[nr+1]>p[pr-1]*p[pr-2]){
            pr-=2; nr+=2;
        }
        else break;
    }
    ll ans=1;
    rep(i,pr) ans=ans*p[i]%MOD;
    rep(i,nr) ans=ans*(ne[i]+MOD)%MOD;
    cout<<(ans+MOD)%MOD<<endl;
}