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: Accepted)

import java.util.*;
import java.io.*;

public class Main {
    public static long MOD = 1000000007L;
    public static void main(String[] args) throws IOException{
        BufferedReader f = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
        StringTokenizer st  = new StringTokenizer(f.readLine());
        int n = Integer.parseInt(st.nextToken());
        int k = Integer.parseInt(st.nextToken());
        ArrayList<Long> pos = new ArrayList<Long>();
        ArrayList<Long> neg = new ArrayList<Long>();
        Long[] arr = new Long[n];
        st = new StringTokenizer(f.readLine());
        long currp = 1L;
        for(int i = 0; i < n; i++){
            long elem = Long.parseLong(st.nextToken());
            arr[i] = elem;
            currp*=elem;
            currp%=MOD;
            currp+=MOD;
            currp%=MOD;
            if(elem > 0) pos.add(elem);
            else neg.add(elem);
        }
        if(k == n){
            out.println(currp);
            out.close();
            System.exit(0);
        }
        Collections.sort(neg);
        long prod = 1L;
        if(pos.size() == 0 && k % 2 == 1){
            for(int i = neg.size()-1; i > neg.size()-1-k; i--){
                prod*=neg.get(i);
                prod%=MOD;
                prod+=MOD;
                prod%=MOD;
            }
            out.println(prod);
            out.close();
            System.exit(0);
        }
        Arrays.sort(arr);
        int a = 0;
        int b = n-1;
        if(k % 2 == 1){
            prod*=arr[b];
            prod%=MOD;
            prod+=MOD;
            prod%=MOD;
            b--;
            k--;
        }

        k/=2;
        for(int i = 0; i < k; i++){
            long left = arr[a]*arr[a+1];
            long right = arr[b]*arr[b-1];
            if(left > right){
                prod*=arr[a];
                prod%=MOD;
                prod+=MOD;
                prod%=MOD;
                prod*=arr[a+1];
                prod%=MOD;
                prod+=MOD;
                prod%=MOD;
                a+=2;
            }else{
                prod*=arr[b];
                prod%=MOD;
                prod+=MOD;
                prod%=MOD;
                prod*=arr[b-1];
                prod%=MOD;
                prod+=MOD;
                prod%=MOD;
                b-=2;
            }
        }




        out.println(prod);
        out.close();
    }
}