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>
//#include <ext/pb_ds/tree_policy.hpp>
//#include <ext/pb_ds/assoc_container.hpp>

using namespace std;
//using namespace __gnu_pbds;

#define ll  long long
#define PR  pair<int, int>

const int N = 3e5 + 10;
const int MOD = 1e9 + 7;
const long long INF = 5e18;
const double PI = 2 * acos(0.0);
//template<typename T> using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;

vector<ll> neg, pos;

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    int n, k;
    cin >> n >> k;
    for (int i = 0; i < n; i++) {
        int x;
        cin >> x;
        if (x < 0) {
            neg.push_back(x);
        } else {
            pos.push_back(x);
        }
    }
    sort(neg.begin(), neg.end());
    sort(pos.begin(), pos.end(), greater<ll>());
    ll ans = 1;
    int i = 0, j = 0, cnt = 0;
    while (i < pos.size() && j < neg.size() && cnt < k) {
        if (cnt + 2 <= k && j + 1 < neg.size() && i + 1 < pos.size()) {
            ll tmp = neg[j] * neg[j + 1];
            ll fmp = pos[i] * pos[i + 1];
            if (pos[i + 1] == 0) {
                fmp = pos[i];
            }
            if (fmp >= tmp) {
                ans *= pos[i];
                ans %= MOD;
                cnt += 1;
                i++;
            } else {
                tmp %= MOD;
                ans *= tmp;
                ans %= MOD;
                cnt += 2;
                j += 2;
            }
        } else {
            ans *= pos[i];
            ans %= MOD;
            cnt++;
            i++;
        }
    }
    if (cnt < k) {
        if (pos.empty()) {
           int baki = k - cnt;
           if (baki % 2) {
                while (cnt < k) {
                    neg.back() += MOD;
                    neg.back() %= MOD;
                    ans *= neg.back();
                    ans %= MOD;
                    neg.pop_back();
                    cnt++;
                }
            } else {
                while (cnt < k) {
                    neg[j] += MOD;
                    neg[j] %= MOD;
                    ans *= neg[j];
                    ans %= MOD;
                    cnt++;
                    j++;
                }
            }
        } else {
            while (cnt < k && i < pos.size()) {
                ans *= pos[i];
                ans %= MOD;
                cnt++;
                i++;
            }
        }
    }
    cout << ans << '\n';
    return 0;
}