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.InputStreamReader;
import java.util.Arrays;
import java.util.Scanner;

public final class Main {

    private static final int MOD = (int) (1e9 + 7);

    static int solve(long[] nums, int k) {
        final int n = nums.length;
        Arrays.sort(nums);
        long product = 1;
        if (nums[n - 1] == 0 && k % 2 != 0) {
            return 0;
        }
        if (nums[n - 1] <= 0 && k % 2 != 0) {
            for (int i = n - 1; i >= n - k; i--) {
                product = (product * nums[i]) % MOD;
            }
            return Math.floorMod(product, MOD);
        }
        final int i = 0;
        final int j = n - 1;
        if (k % 2 != 0) {
            return Math.floorMod(Math.max(get(nums, k - 1, nums[j], i, j - 1),
                                          get(nums, k - 1, nums[i], i + 1, j)), MOD);
        }
        k >>= 1;
        return Math.floorMod(get(nums, k, product, i, j), MOD);
    }

    private static long get(long[] nums, int k, long product, int i, int j) {
        for (int itr = 0; itr < k && i < j; itr++) {
            final long left = (nums[i] * nums[i + 1]) % MOD;
            final long right = (nums[j] * nums[j - 1]) % MOD;
            if (left > right) {
                product = (product * left) % MOD;
                i += 2;
            } else {
                product = (product * right) % MOD;
                j -= 2;
            }
        }
        return product;
    }

    public static void main(String[] args) {
        final Scanner in = new Scanner(new BufferedReader(new InputStreamReader(System.in)));
        final int n = in.nextInt();
        final int k = in.nextInt();
        in.nextLine();
        final long[] arr = new long[n];
        for (int i = 0; i < n; i++) {
            arr[i] = in.nextLong();
        }
        in.nextLine();
        System.out.println(solve(arr, k));
    }
}