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)

//#define _GLIBCXX_DEBUG
#include <bits/stdc++.h>
using namespace std;
using ll=long long;
using ld=long double;
using ch=char;
typedef pair<ll,ll> P;
typedef vector<ll> vl;
typedef vector<vl> vvl;
typedef vector<P> vP;
typedef vector<ch> vc;
typedef vector<vc> vvc;
#define FOR(i,a,b) for(ll i=a;i<b;i++)
#define rep(i,n) FOR(i,0,n)
#define ROF(i,a,b) for(ll i=a;i>=b;i--)
#define per(i,a) ROF(i,a,0)
#define pb push_back
const ll MOD=1000000007;
const ll MOD2=998244353;
const ld PI=acos(-1);
const ll INF=1e18;
string abc="abcdefghijklmnopqrstuvwxyz";
string ABC="ABCDEFGHIJKLMNOPQRSTUVWXYZ";
struct edge{ll to,cost;};
struct edge2{ll from,to,cost;};


int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    ll n,k;
    cin >> n >> k;
    vl p,m;
    rep(i,n){
        ll x;
        cin >> x;
        if(x>=0){
            p.pb(x);
        }
        else{
            m.pb(-x);
        }
    }
    if(n==k){
        ll ans=1;
        rep(i,p.size()){
            ans*=p[i];
            ans%=MOD;
        }
        rep(i,m.size()){
            ans*=-m[i];
            ans%=MOD;
        }
        cout << (MOD+ans)%MOD << endl;
        return 0;
    }
    sort(p.begin(),p.end());
    reverse(p.begin(),p.end());
    sort(m.begin(),m.end());
    reverse(m.begin(),m.end());
    ll msz=m.size(),psz=p.size();
    ll z=min(k,psz);
    if(psz==0){
        ll ans=1;
        if(k%2==0){
            rep(i,k){
                ans*=m[i];
                ans%=MOD;
            }
            cout << ans << endl;
            return 0;
        }
        else{
            rep(i,k){
                ans*=m[msz-1-i];
                ans%=MOD;
            }
            cout << (MOD-ans)%MOD << endl;
            return 0;
        }
    }
    if(psz==1&&k%2==0){
        ll ans=1;
        rep(i,k){
            ans*=m[i];
            ans%=MOD;
        }
        cout << ans << endl;
        return 0;
    }
    if((z-k)%2==1){
        z--;
    }
    ll f=0;
    while(z-f-2>=0&&f+1<msz&&m[f]*m[f+1]>p[z-f-1]*p[z-f-2]){
        f+=2;
    }
    ll ans=1;
    rep(i,f){
        ans*=m[i];
        ans%=MOD;
    }
    rep(i,(k-f)){
        ans*=p[i];
        ans%=MOD;
    }
    cout << ans << endl;
}