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: Accepted)

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#define all(x) (x).begin(), (x).end()

using namespace __gnu_pbds;
using namespace std;

using ull = unsigned long long;
using ll = long long;
using ld = long double;
template <typename T> using ordered_set = tree<T, null_type, less<>, rb_tree_tag, tree_order_statistics_node_update>;

const int mod = 1e9 + 7;
const int inf = INT_MAX;
const int N = 2e5 + 5;

int n, k, a[N];

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

    cin >> n >> k;
    for (int i = 0; i < n; i++) cin >> a[i];

    sort(a, a + n);
    if (a[n - 1] == 0 and (k & 1)) {
        cout << "0\n";
        return 0;
    }

    if (a[n - 1] < 0 and (k & 1)) {
        ll prod = 1;
        for (int i = n - k; i < n; i++)
            prod = prod * a[i] % mod;
        prod = (prod + mod) % mod;
        cout << prod << "\n";
        return 0;
    }

    ll prod = 1;
    if (a[n - 1] > 0 and (k & 1)) {
        prod = a[n - 1];
        n--;
    }

    int i = n - 1, j = 1;
    k--;

    while (n - 1 - i + j - 1 < k) {
        if (a[i] * (ll)a[i - 1] > a[j] * (ll)a[j - 1])
            prod = prod * a[i] % mod * a[i - 1] % mod, i -= 2;
        else prod = prod * a[j] % mod * a[j - 1] % mod, j += 2;
    }

    prod = (prod + mod) % mod;
    cout << prod << "\n";
    return 0;
}