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 rep(i,n) for(int i=0;i<(int)(n);i++)
#define FOR(i,n,m) for(int i=(int)(n); i<=(int)(m); i++)
#define RFOR(i,n,m) for(int i=(int)(n); i>=(int)(m); i--)
#define ITR(x,c) for(__typeof(c.begin()) x=c.begin();x!=c.end();x++)
#define RITR(x,c) for(__typeof(c.rbegin()) x=c.rbegin();x!=c.rend();x++)
#define setp(n) fixed << setprecision(n)

template<class T> bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T> bool chmin(T &a, const T &b) { if (a>b) { a=b; return 1; } return 0; }

#define ll long long
#define vll vector<ll>
#define vi vector<int>
#define pll pair<ll,ll>
#define pi pair<int,int>

#define all(a) (a.begin()),(a.end())
#define rall(a) (a.rbegin()),(a.rend())
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define ins insert

using namespace std;

//-------------------------------------------------
//--ModInt
//-------------------------------------------------
const ::std::uint_fast64_t MOD = 1e9+7;

class mint
{
private:
    using value_type = ::std::uint_fast64_t;
    value_type n;
public:
    mint():n(0){}
    mint(::std::int_fast64_t _n):n(_n<0 ? MOD-(-_n)%MOD : _n%MOD){}
    mint(const mint &m):n(m.n){}

    friend ::std::ostream& operator<<(::std::ostream &os, const mint &a){
		return os << a.n;
	}
    friend ::std::istream& operator>>(::std::istream &is, mint &a){
        value_type temp; is>>temp;
        a = mint(temp);
        return is;
    }

    mint& operator+=(const mint &m){n+=m.n; n=(n<MOD)?n:n-MOD; return *this;}
    mint& operator-=(const mint &m){n+=MOD-m.n; n=(n<MOD)?n:n-MOD; return *this;}
    mint& operator*=(const mint &m){n=n*m.n%MOD; return *this;}
    mint& operator/=(const mint &m){return *this*=m.inv();}
    mint& operator++(){return *this+=1;}
    mint& operator--(){return *this-=1;}

    mint operator+(const mint &m) const {return mint(*this)+=m;}
    mint operator-(const mint &m) const {return mint(*this)-=m;}
    mint operator*(const mint &m) const {return mint(*this)*=m;}
    mint operator/(const mint &m) const {return mint(*this)/=m;}
    mint operator++(int){mint t(*this); *this+=1; return t;}
    mint operator--(int){mint t(*this); *this-=1; return t;}

    bool operator==(const mint &m) const {return n==m.n;}
    bool operator!=(const mint &m) const {return n!=m.n;}
    mint operator-() const {return mint(MOD-n);}

    mint pow(value_type b) const {
        mint ret(1), m(*this);
        while(b){
            if (b & 1) ret*=m;
            m*=m;
            b>>=1;
        }
        return ret;
    }

    mint inv() const {return pow(MOD-2);}
};

//-------------------------------------------------

int main(void)
{
    cin.tie(0);
    ios::sync_with_stdio(false);
    int N,K; cin>>N>>K;
    vll a(N);
    rep(i,N) cin>>a[i];
    vll p,n;
    bool ok=false;
    rep(i,N){
        if (a[i]>0){
            p.pb(a[i]);
        }else if(a[i]<0){
            n.pb(a[i]);
        }else{
            ok=true;
        }
    }
    sort(all(n));
    sort(rall(p));
    int M = n.size();
    int L = p.size();
    vector<mint> nmul(M+1,1), pmul(L+1,1);
    rep(i,M) nmul[i+1]=nmul[i]*n[i];
    rep(i,L) pmul[i+1]=pmul[i]*p[i];

    // rep(i,M+1) cout<<nmul[i]<<" "; cout<<endl;
    // rep(i,L+1) cout<<pmul[i]<<" "; cout<<endl;

    mint nMax, pMax;
    bool pok=false;
    for(int x=0; x<=M; x+=2){
        if (!(0<=K-x && K-x<=L)) continue;
        if (!pok){
            pMax = pmul[K-x]*nmul[x];
            pok = true;
        }else{
            int y = K-x;
            ll A = n[x-1]*n[x-2];
            ll B = p[y]*p[y+1];
            if (B>=A) break;
            pMax = pmul[y]*nmul[x];
        }
    }

    reverse(all(n));
    reverse(all(p));
    nmul[0] = 1;
    pmul[0] = 1;
    rep(i,M) nmul[i+1]=nmul[i]*n[i];
    rep(i,L) pmul[i+1]=pmul[i]*p[i];

    bool nok=false;
    for(int x=1; x<=M; x+=2){
        if (!(0<=K-x && K-x<=L)) continue;
        if (!nok){
            nMax = pmul[K-x]*nmul[x];
            nok = true;
        }else{
            int y = K-x;
            ll A = n[x-1]*n[x-2];
            ll B = p[y]*p[y+1];
            if (B<=A) break;
            nMax = pmul[y]*nmul[x];
        }
    }

    if (!pok && ok){
        cout<<"0\n";
    }else if(!pok){
        cout<<nMax<<endl;
    }else{
        cout<<pMax<<endl;
    }

    return 0;
}