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)

#include <bits/stdc++.h>

using namespace std;
using ll = long long;

const long long MOD = 1000000007;
int main() {
    ll N = 0;
    cin >> N;
    ll K = 0;
    cin >> K;
    ll countp = 0, countz = 0, countn = 0;
    vector<ll> A(N,0);
    for(ll i = 0; i < N; i++) {
    cin >> A.at(i);
    if (A.at(i) > 0)countp++;
    else if (A.at(i) == 0)countz++;
    else countn++;
    }

    sort(A.begin(),A.end());
    ll ans = 1;
    ll getp;
    ll getn;
    
    if (countp + countz == 0)
    {
        for (size_t i = 0; i < K; i++)
        {
            ans = ((ans*A.at(N-1-i)%MOD)+MOD)%MOD;
        }
        cout << ans << endl;
        return 0;
    }
    

    if (K%2)
    {
        getn = 0;
        getp = 0;
        ans = A.at(N-1);
        for (size_t i = 0; i < K-1; i += 2)
        {
            if (A.at(N-2-getp)*A.at(N-3-getp) >= A.at(getn)*A.at(getn+1))
            {
                ans = ((ans*A.at(N-2-getp)%MOD)+MOD)%MOD;
                ans = ((ans*A.at(N-3-getp)%MOD)+MOD)%MOD;
                getp += 2;
            }
            else
            {
                ans = ((ans*A.at(getn)%MOD)+MOD)%MOD;
                ans = ((ans*A.at(1+getn)%MOD)+MOD)%MOD;
                getn += 2;
            }
            
        }
        cout << ans << endl;
        
        
    }else
    {
        getn = 0;
        getp = 0;
        for (size_t i = 0; i < K; i += 2)
        {
            //cout << ans << endl;
            if (A.at(N-2-getp)*A.at(N-1-getp) >= A.at(getn)*A.at(getn+1))
            {
                ans = ((ans*A.at(N-2-getp)%MOD)+MOD)%MOD;
                ans = ((ans*A.at(N-1-getp)%MOD)+MOD)%MOD;
                getp += 2;
            }
            else
            {
                ans = ((ans*A.at(getn)%MOD)+MOD)%MOD;
                //cout << ans << endl;
                //cout << ans*A.at(1+getn) << endl;
                ans = ((ans*A.at(1+getn)%MOD)+MOD)%MOD;
                getn += 2;
            }
            
        }
        cout << ans << endl;
    }
    
    



    return 0;
}