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.Arrays;
import java.util.Comparator;
import java.util.Scanner;

public class Main {
    static final long MOD = 1_000_000_007;
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        int k = in.nextInt();
        Long[] a = new Long[n];

        boolean allNegative = true;
        for (int i = 0; i < n; i++) {
            a[i] = in.nextLong();
            if (a[i] >= 0) {
                allNegative = false;;
            }
        }

        Arrays.sort(a, Comparator.comparingLong(v -> Math.abs(v)));

        if (allNegative) {
            long ans = 1;
            for (int i = 0; i < k; i++) {
                ans = modmul(ans, a[i], MOD);
            }

            if (ans < 0) {
                ans += MOD;
            }
            System.out.println(ans);
            return;
        }

        long ans = 1;
        int index = n - 1;
        for (int count = 0; count < k - 1 ; count++) {
            ans = modmul(ans, a[index], MOD);
            index--;
        }

        if (k % 2 == 0) {
            ans = modmul(ans, a[index], MOD);
        } else {
            for (int i = index; i >= 0; i--) {
                if (a[i] >= 0) {
                    ans = modmul(ans, a[i], MOD);
                }
            }
        }

        if (ans < 0) {
            ans += MOD;
        }
        System.out.println(ans);
    }

    static long modmul(long a, long b, long mod) {
        return (a * b) % mod;
    }
}