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 ll mod = 1e9 + 7;



int main()
{
    int n, k;
    cin >> n >> k;
    vector<ll> se, fu;
    int zero = 0;
    for(int i = 0; i < n; i++) {
        int t; cin >> t;
        if(t > 0)se.emplace_back(t);
        if(t <= 0)fu.emplace_back(-t);
        if(t == 0)zero++;

    }

    sort(se.begin(), se.end());
    sort(fu.begin(), fu.end());

    if(n - zero < k) {
        cout << 0 << endl;
        return 0;
    }

    //負 or ゼロになる
    if((int)fu.size() == n) {
        ll ans = 1;
        for(int i = 0; i < k; i++)ans *= -fu[i], ans %= mod;
        cout << (ans % mod + mod) % mod << endl;
        return 0;
    }

    ll ans = 1;
    if(k % 2) {
        ans = se.back();
        se.pop_back();
    }
    k /= 2;

    for(int i = 0; i < k; i++) {
        ll pl = (se.size() >= 2 ? se.back() * *(se.end() - 2) : 1);
        ll mi = (fu.size() >= 2 ? fu.back() * *(fu.end() - 2) : 1);
    
        if(pl >= mi && se.size() >= 2) {
            ans *= pl % mod;
            ans %= mod;
            se.pop_back(), se.pop_back();
        } else if(mi > pl && fu.size() >= 2){
            ans *= mi % mod;
            ans %= mod;
            fu.pop_back(), fu.pop_back();
        } else {
            while(true);
            // cerr << fu.size() << " : " << se.size() << endl;
        }
    }
    cout << ans << endl;
}