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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.Comparator;
import java.util.*;
public class Main {
	static long mod=1000000007;
	 static long maxProduct(long A[], int n, int k) 
	    { 
	        Arrays.sort(A); 
	        long product = 1; 
	        if (A[n - 1] == 0 && k % 2 != 0) 
	            return 0; 
	        if (A[n - 1] <= 0 && k % 2 != 0) { 
	            for (int i = n - 1; i >= n - k; i--) 
	            {
	            	product=(product*A[i])%mod; 
	            }
	            return (product+mod)%mod; 
	        } 
	        int i = 0; 
	        int j = n - 1; 
	        if (k % 2 != 0) { 
	            product=(product*A[j])%mod; 
	            j--; 
	            k--; 
	        }
	        k >>= 1; 
	        for (int itr = 0; itr < k; itr++) { 
	            long left_product = (A[i] * A[i + 1])%mod; 
	            long right_product = (A[j] * A[j - 1])%mod; 
	            if (left_product > right_product) { 
	                product=(product*left_product)%mod; 
	                i += 2; 
	            } 
	            else { 
	                product=(product*right_product)%mod; 
	                j -= 2; 
	            } 
	        } 
	        return (product+mod)%mod; 
	} 
	public static void main(String[] args) throws IOException {
		// TODO Auto-generated method stub
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		String[] cmd=br.readLine().split(" ");
		int n=Integer.valueOf(cmd[0]);
		int k=Integer.valueOf(cmd[1]);
		long[] arr=new long[n];
		cmd=br.readLine().split(" ");
		for(int i=0;i<n;i++)
		{
			arr[i]=Long.valueOf(cmd[i]);
		}
		System.out.println(maxProduct(arr,n,k));
	}
}