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 = 1e9 + 7;

int main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    
    int n, k;
    cin >> n >> k;

    vector<long long> neg, pos;
    for (int i = 0; i < n; i++){
        long long a;
        cin >> a;

        if (a < 0)
            neg.push_back(a);
        else
            pos.push_back(a);
    }

    if (k % 2 && (int) pos.size() == 0){
        sort(neg.begin(), neg.end(), greater<long long>());

        long long ans = 1;
        for (int i = 0; i < k; i++)
            ans = (ans * neg[i]) % MOD;
        
        while (ans < 0)
            ans += MOD;

        cout << ans % MOD << '\n';
    }
    else{
        sort(pos.begin(), pos.end());
        sort(neg.begin(), neg.end());

        long long ans = 1;
        if (k % 2){
            ans *= pos.back();
            pos.pop_back();
            k -= 1;
        }

        reverse(pos.begin(), pos.end());
        
        vector<long long> mul;
        for (int i = 0; i < (int) pos.size() && i + 1 < (int) pos.size(); i += 2)
            mul.push_back((pos[i] * pos[i + 1]) % MOD);

        for (int i = 0; i < (int) neg.size() && i + 1 < (int) neg.size(); i += 2)
            mul.push_back((neg[i] * neg[i + 1]) % MOD);

        sort(mul.begin(), mul.end(), greater<long long>());

        for (int i = 0; i < k / 2; i++)
            ans = (ans * mul[i]) % MOD;

        cout << ans << '\n';
    }

    return 0;
}