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++ (Clang 10.0.0) (judge verdict: Wrong Answer)

#include <iostream>
#include <queue>
#include <vector>
#include <algorithm>

#define int long long
#define MOD (int)(1e9+7)

using namespace std;

signed main(){
    int N, K;
    cin >> N >> K;
    int n_zero = 0;
    priority_queue<int> p,m;
    for(int i=0;i<N;i++){
        int x;
        cin >> x;
        if(x == 0){
            n_zero++;
        }else if(x>0){
            p.push(x);
        }else{
            m.push(-x);
        }
    }

    if(N-n_zero<K){
        cout << 0 << endl;
        return 0;
    }

    // possible to make positive ans
    if(m.size()/2 + p.size() >= K){
        int ans = 1;
        if(K%2==1 && !p.empty()){
            ans = p.top();
            p.pop();
            K--;
        }

        priority_queue<int> mul;
        while(p.size() >= 2){
            int x = p.top();
            p.pop();
            x *= p.top();
            p.pop();
            x %= MOD;
            mul.push(x);
        }
        while (m.size() >= 2)
        {
            int x = m.top();
            m.pop();
            x *= m.top();
            m.pop();
            x %= MOD;
            mul.push(x);
        }
        for(int i=0;i<K/2;i++){
            ans *= mul.top();
            mul.pop();
            ans %= MOD;
        }
        cout << ans << endl;
    }else{
        int ans = 1;
        vector<int> v;
        while (!p.empty())
        {
            int x = p.top();
            p.pop();
            v.push_back(x);
        }
        while (!m.empty())
        {
            int x = m.top();
            m.pop();
            v.push_back(x);
        }
        sort(v.begin(), v.end());
        for(int i=0;i<K;i++){
            ans *= v[i];
            ans %= MOD;
        }
        cout << (-ans + MOD) % MOD << endl;
    }
    return 0;
}