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: Time Limit Exceeded)

#include <bits/stdc++.h>
using namespace std;
typedef long long int ll;
typedef long double ld;
#define MOD 1000000007
#define ALL(obj) (obj).begin(),(obj).end()
template<class T>inline bool chmax(T& a,T b){if (a<b){a=b;return 1;}return 0;}
template<class T>inline bool chmin(T& a,T b){if (a>b){a=b;return 1;}return 0;}
const long long INF = 1LL << 60;
bool pairCompare(const pair<double,ll>& firstElof, const pair<double,ll>& secondElof){
    return firstElof.first < secondElof.first;
}
bool pairCompareSecond(const pair<double,ll>& firstElof, const pair<double,ll>& secondElof){
    return firstElof.second < secondElof.second;
}
//四方向への移動ベクトル
const int dx[4] = {1, 0, -1, 0};
const int dy[4] = {0, 1, 0, -1};

struct edge{//グラフに使うヤツ
    ll from,to,cost;
};

typedef vector<vector<edge> > G;
ll gcd(ll a,ll b){
    if (a%b==0)return(b);
    else return(gcd(b,a%b));
}



int main() {
    ll n,k;
    cin >> n >> k;
    ll a[n];
    for (ll i=0;i<n;i++)cin >> a[i];
    vector<ll> p;
    vector<ll> m;
    for (ll i=0;i<n;i++){
        if (a[i]>=0)p.push_back(a[i]);
        else m.push_back(a[i]);
    }
    if (p.size())sort(ALL(p));
    if (m.size())sort(ALL(m),greater<ll>());
    vector<ll> ansvec;
    ll ans=1;
    ll x=p.size()-1,y=m.size()-1;
    ll c=0;
    if (m.size()){
        while(true);
    }
    if (p.size()==0){
        for (ll i=0;i<k;i++){
            ans*=m[i];
        }
        cout << 1000000007+ans << endl;
        return 0;
    }
    for (ll i=0;i<k;i++){
        if (x>=0 and y>=0){
            if (p[x]>abs(m[y])){
                ansvec.push_back(p[x]);
                x--;
            }else{
                ansvec.push_back(m[y]);
                y--;
                c++;
            }
        }else if(x>=0){
            ansvec.push_back(p[x]);
            x--;
        }else if (y>=0){
            ansvec.push_back(m[y]);
            y--;
            c++;
        }
    }
    if(c%2==1 and x>=0){
        sort(ALL(ansvec));
        ll f=-INF;
        for (ll i=0;i<n;i++){
            if (ansvec[i]<0){
                f=max(ansvec[i],f);
            }
        }
        ansvec.push_back(p[x]);
        for (ll i=0;i<ansvec.size();i++){
            if (ansvec[i]==f)continue;
            ans*=ansvec[i];
            ans%=MOD;
        }
        cout << ans << endl;
    }else{
        for (ll i=0;i<ansvec.size();i++){
            ans*=ansvec[i];
            ans%=MOD;
        }
        cout << ans << endl;
    }
    return 0;
}