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 Java (OpenJDK 11.0.6) (judge verdict: Wrong Answer)

import java.math.BigDecimal;
import java.util.*;

public class Main{
    public static void main(String[] args){
        Scanner sc = new Scanner(System.in);

        int n = sc.nextInt();
        int k = sc.nextInt();

        int MOD = 1000000007;

        int[] a = new int[n];
        PriorityQueue<Integer> pos = new PriorityQueue<>((x, y) -> y - x);
        PriorityQueue<Integer> neg = new PriorityQueue<>((x, y) -> x - y);
        PriorityQueue<Integer> neg2 = new PriorityQueue<>((x, y) -> y - x);

        for(int i = 0; i < n; i++){
            a[i] = sc.nextInt();
            if(a[i] > 0){
                pos.add(a[i]);
            }else{
                neg.add(a[i]);
                neg2.add(a[i]);
            }
        }

        long ans = 1;

        if(pos.isEmpty() && k % 2 == 1){
            for(int i = 0; i < k; i++){
                ans *= neg2.poll();
                ans %= MOD;
            }
            System.out.println(MOD + ans);
            return;
        }

        long p = 1;
        long po = 1;
        if(pos.size() > 0){
            p = pos.poll();
            if(pos.size() > 0){
                po = p * pos.peek();
            }else{
                po = p;
            }
        }
        long ne = 1;
        if(neg.size() > 1){
            ne = neg.poll() * neg.poll();
        }

        for(int i = 0; i < k; i++){
            if(po > ne){
                ans *= p;
                ans %= MOD;
                if(pos.size() > 0){
                    p = pos.poll();
                    if(pos.size() > 0){
                        po = p * pos.peek();
                    }else{
                        po = p;
                    }
                }
            }else{
                ans *= ne;
                ans %= MOD;
                if(neg.size() > 1){
                    ne = neg.poll() * neg.poll();
                }
                i++;
            }
            //System.out.println(ans);
        }

        System.out.println(ans % 1000000007);

    }
}