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.util.ArrayList;
import java.util.Collections;
import java.util.List;
import java.util.Scanner;

public class Main {
	
	static int mod = 1000000007;
	
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		
		int n = Integer.parseInt(sc.next());
		int k = Integer.parseInt(sc.next());
		
		List <Integer> po = new ArrayList<Integer>();
		List <Integer> ne = new ArrayList<Integer>();
		
		for(int i = 0 ; i < n ;i++) {
			int a = Integer.parseInt(sc.next());
			if(a >= 0) {
				po.add(a);
			}
			else {
				ne.add(a);
			}
		}
		Collections.sort(po);
		Collections.sort(ne);
				
		long []pol = new long [po.size()+1];
		long []nel = new long [ne.size()+1];
		long []nel2 = new long [ne.size()+1];
		nel[0] = 1;
		pol[0] = 1;
		nel2[0] = 1;
		
		for(int i = 0 ; i < po.size() ;i++) {
			pol[i+1] = (long)(pol[i] * po.get(po.size()-i-1))%mod;
		}
		for(int i = 0 ; i < ne.size() ; i++) {
			nel[i+1] = (long)(nel[i]* ne.get(i))%mod;
			nel2[i+1] = (long)(nel2[i]*ne.get(ne.size()-i-1))%mod;
		}
		long ans = -1000000000000000000L;
		
		if(po.size()==0) {
			System.out.println(mod+nel2[k]);
			return;
		}
		for(int i = 0 ; i <= Math.min(k, ne.size()) ; i+=2) {
			int t = k - i;
			if(t > po.size()) {
				continue;
			}
			long tmp = (long)(nel[i]*pol[t])%mod;
			ans = Math.max(tmp,ans);
		}
		
		System.out.println(ans);
	}
}