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;
const long long MOD = 1000000007;

int main (){
    long long N = 0;
    cin >> N;
    vector<long long> A(N,0);
    //vector<long long> B(N,0);
    //vector<long long> C(N,0);

    long long K = 0;
    cin >> K;
    for(long long i = 0; i < N; i++) {
    cin >> A.at(i);
    //B.at(i) = abs(A.at(i));

    }
    sort(A.begin(),A.end());
    reverse(A.begin(),A.end());
    long long p = 0;
    long long n = 0;
    long long ans = 0;
    if(K%2) {
        ans = A.at(0);
        p = 1;
        for (size_t i = 0; i < K-1; i++)
        {
            if (A.at(p)*A.at(p+1) >= A.at(N-1-n)*A.at(N-2-n))
            {
                ans = ans*A.at(p)*A.at(p+1)%MOD;


                p += 2;
            }else
            {
                ans = ans* A.at(N-1-n)*A.at(N-2-n)%MOD;
                /* code */
                n += 2;
            }
            i++;
            
        }
        
    } else {
        ans = 1;
        for (size_t i = 0; i < K; i++)
        {
            if (A.at(p)*A.at(p+1) >= A.at(N-1-n)*A.at(N-2-n))
            {
                ans = ans *A.at(p)*A.at(p+1)%MOD;


                p += 2;
            }else
            {
                ans = ans * A.at(N-1-n)*A.at(N-2-n)%MOD;
                /* code */
                n += 2;
            }
            i++;
            
            
        }
    }
    cout << ans << endl;

    
}