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: Wrong Answer)

//#define _GLIBCXX_DEBUG
#include <bits/stdc++.h>
using namespace std;
#define all(v) v.begin(), v.end()
using in = int64_t;
#define rep(i, n) for (in i = 0; i < (in)(n); i++)
#define REP(i,a,b) for(in i=(in)(a);i<(in)(b);i++)
#define Yes cout<<"Yes"<<endl
#define No cout<<"No"<<endl
#define yes cout<<"yes"<<endl
#define no cout<<"no"<<endl
const double PI=3.14159265358979323846;
const in MOD = 1000000007;
const in INF=1e18+7;
const in inf=1e9+7;
using P=pair<in,in>;
vector<in> dx={0,1,-1,0};
vector<in> dy={1,0,0,-1};


int main(){
ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
cout << fixed << setprecision(10);
    in n,k;
    cin>>n>>k;
    vector<in> a(n);
    rep(i,n) cin>>a[i];
    vector<in> sei(0),hu(0);
    // in cnt_zero=0;
    rep(i,n){
        if(a[i]>0) sei.emplace_back(a[i]);
        else hu.emplace_back(a[i]);
    }
    sort(all(sei));
    sort(all(hu));
    reverse(all(sei));

    if(n==k){
        in ans=1ll;
        rep(i,n){
            ans *=a[i];
            ans %=MOD;
        }
        if(ans<MOD) ans +=MOD;
        cout<<ans<<endl;
        return 0;
    }
    if(sei.empty()){
        if(k%2==0){
            in ans=1ll;
            rep(i,k){
                ans *= hu[i];
                ans %= MOD;
            }
            cout<<ans<<endl;
            return 0;
        }
        else{
            in ans=1ll;
            rep(i,k){
                ans *= hu[n-1-i];
                ans %=MOD;
            }
            if(ans<0) ans +=MOD;
            cout<<ans<<endl;
            return 0;
        }
    }

    in ans=1ll,sei_num=0,hu_num=0;
    if(k%2==1){
        ans *=sei[0];
        sei_num ++;
    }

    for(in i=0;i<k/2;i++){
        in kou_sei=0ll,kou_hu=0ll;
        if(sei_num-1ll<(in)sei.size() && !sei.empty()) kou_sei=sei[sei_num]*sei[sei_num+1];
        if(hu_num-1ll<(in)hu.size() && !hu.empty()) kou_hu =hu[hu_num]*hu[hu_num+1];

        if(kou_hu>kou_sei){
            kou_hu %=MOD;
            ans *=kou_hu;
            ans %=MOD;
            hu_num +=2;
        }
        else{
            kou_sei %=MOD;
            ans *=kou_sei;
            ans %=MOD;
            sei_num +=2;
        }
    }
    cout<<ans<<endl;
    //cout<<hu.size()<<" "<<sei.size()<<endl;
}/*
./problem.exe
*/