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.*;

public class Main {
	static int r = 1000000000 + 7;

	public static void main(String args[]) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt(), k = sc.nextInt();
		int a[] = new int[n];
		for (int i = 0; i < n; ++i)
			a[i] = sc.nextInt();
		sc.close();
		Arrays.sort(a);
		long ans = 1, minus = 1;
		for (int i = 0; i < k; ++i) {
			ans *= (long) a[n - i - 1];
			ans %= r;
		}
		if ((long) a[0] * (long) a[n - 1] < 0) {
			long tmp = ans;
			for (int i = 2; i <= k; i += 2) {
				if (a[i - 2] != 0) {
					minus *= (long) a[i - 2];
					minus %= r;
				}
				if (a[i - 1] != 0) {
					minus *= (long) a[i - 1];
					minus %= r;
				}
				minus %= r;
				if (a[n - k + i - 2] != 0) {
					tmp /= (long) a[n - k + i - 2];
					tmp %= r;
				}
				if (a[n - k + i - 1] != 0) {
					tmp /= (long) a[n - k + i - 1];
					tmp %= r;
				}
				tmp *= minus;
				tmp %= r;
				ans = Math.max(tmp, ans);
			}
		}
		if (ans < 0) {
			ans += r;
		}
		System.out.println(ans);
	}
}