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>
#define rep(i, n) for (int i = 0; i < (n); ++i)
using namespace std;
using ll = long long;
using P = pair<int, int>;

const int mod = 1000000007;

int main() {
    ll N, K;
    cin >> N >> K;
    vector<ll> A(N);
    rep(i, N) cin >> A[i];

    ll ans = 1;
    if (N == K)
        rep(i, K) ans = (ans * A[i]) % mod;
    else {
        vector<ll> plus, minus;
        rep(i, N) {
            if (A[i] < 0)
                minus.push_back(A[i]);
            else
                plus.push_back(A[i]);
        }

        if (plus.size() == 0 && K % 2 == 1) {
            sort(minus.begin(), minus.end(), greater<>());
            rep(i, K) ans = (ans * minus[i]) % mod;
        } else {
            vector<pair<ll, pair<ll, ll>>> product;
            vector<ll> rest;
            if (!plus.empty()) {
                sort(plus.begin(), plus.end());
                for (int i = 0; i < plus.size(); i += 2) {
                    if (i == plus.size() - 1) {
                        rest.push_back(plus[i]);
                        break;
                    }
                    product.push_back(
                        make_pair(plus[i] * plus[i + 1],
                                  make_pair(plus[i], plus[i + 1])));
                }
            }
            if (!minus.empty()) {
                sort(minus.begin(), minus.end());
                for (int i = 0; i < minus.size(); i += 2) {
                    if (i == minus.size() - 1) {
                        rest.push_back(minus[i]);
                        break;
                    }
                    product.push_back(
                        make_pair(minus[i] * minus[i + 1],
                                  make_pair(minus[i], minus[i + 1])));
                }
            }

            sort(product.begin(), product.end(), greater<>());
            for (int i = 0; i * 2 < K; ++i) {
                if (i * 2 == K - 1) {
                    for (int j = i; j < product.size(); ++j) {
                        rest.push_back(product[j].second.first);
                        rest.push_back(product[j].second.second);
                    }
                    sort(rest.begin(), rest.end(), greater<>());
                    ans = (ans * rest[0]) % mod;
                    break;
                }
                ans = (ans * product[i].first) % mod;
            }
        }
    }

    if (ans < 0) ans += mod;
    cout << ans << endl;
}