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;
using ld = long double;
using uint = unsigned int;
using ull = unsigned long long;
using pii = pair<int, int>;

/* [💭] [💡] [🎈] */

const ll mod = 1000000007;

int main() {
    cin.tie(nullptr);
    cout.tie(nullptr);
    ios_base::sync_with_stdio(false);

    int n, k;
    cin >> n >> k;

    priority_queue<int> pos, neg;
    for (int i = 0; i < n; i++) {
        int x;
        cin >> x;
        if (x < 0) neg.emplace(-x);
        if (x > 0) pos.emplace(x);
    }

    if (k > (pos.size() + neg.size())) {
        cout << 0;
        return 0;
    }

    if (!pos.size() && (k & 1)) {
        if (n > (pos.size() + neg.size())) {
            cout << 0;
            return 0;
        }

        ll p = 1;
        priority_queue<int> revneg;
        while (neg.size()) {
            revneg.emplace(-neg.top());
            neg.pop();
        }
        while (k--) {
            p *= (mod + revneg.top());
            revneg.pop();
            p %= mod;
        }

        cout << p;
        return 0;
    }

    if (pos.size() + neg.size() == k && (neg.size() & 1) && n > (pos.size() + neg.size())) {
        cout << 0;
        return 0;
    }

    ll p = 1;
    int l = k;
    while (l >= 1) {
        if (pos.size() >= 2 && neg.size() >= 2 && l >= 2) {
            ll p1 = pos.top();
            pos.pop();
            ll p2 = pos.top();
            pos.emplace(p1);
            ll n1 = neg.top();
            neg.pop();
            ll n2 = neg.top();
            neg.emplace(n1);
            if (p1 * p2 > n1 * n2) {
                p = ((p * p1) % mod * p2) % mod;
            } else {
                p = ((p * n1) % mod * n2) % mod;
            }
            l -= 2;
        } else if (neg.size() >= 2 && l >= 2) {
            ll n1 = neg.top();
            neg.pop();
            ll n2 = neg.top();
            neg.emplace(n1);
            p = ((p * n1) % mod * n2) % mod;
            l -= 2;
        } else if (pos.size()) {
            p = (p * pos.top()) % mod;
            pos.pop();
            l--;
        } else {
            p = (-p * neg.top()) % mod;
            neg.pop();
            l--;
        }
    }

    p = ((p % mod) + mod) % mod;
    cout << p;

    return 0;
}