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 {

  private static final long MOD = 1_000_000_007;

  public static void main(String[] args) {
    int n = scanner.nextInt();
    int k = scanner.nextInt();
    long[] A = new long[n];
    for (int i = 0; i < n; i++) {
      A[i] = scanner.nextInt();
    }
    List<Long> pos = new ArrayList<>();
    List<Long> neg = new ArrayList<>();
    int zero = 0;
    for (long i : A) {
      if (i > 0) {
        pos.add(i);
      } else if (i < 0) {
        neg.add(i);
      } else {
        zero++;
      }
    }
    if (pos.size() + neg.size() < k || (n == k && zero > 0)) {
      System.out.println(0);
      return;
    }
    if (n == k) {
      long res = 1;
      for (long i : A) {
        res *= i;
        res %= MOD;
      }
      System.out.println(Math.floorMod(res, MOD));
      return;
    }
    Arrays.sort(A);
    long res = 1;
    int j = A.length - 1;
    int i = 0;
    if (A[j] < 0 || k == 1) {
      for (; k > 0; k--) {
        res *= A[j--];
        res %= MOD;
      }
      System.out.println(Math.floorMod(res, MOD));
      return;
    }
    if (k % 2 != 0) {
      // res *= A[j--];
      // res %= MOD;
      res = mul(res, A[j--]);
    }
    for (; k > 0; k -= 2) {
      long lft = A[i] * A[i + 1];
      long rgt = A[j] * A[j - 1];
      if (lft > rgt) {
        // lft %= MOD;
        // res *= lft;
        // res %= MOD;
        res = mul(res, lft);
        i += 2;
      } else {
        // rgt %= MOD;
        // res *= rgt;
        // res %= MOD;
        res = mul(res, rgt);
        j -= 2;
      }
    }
    System.out.println(Math.floorMod(res, MOD));
  }

  private static long mul(long a, long b) {
    return (((a + MOD) % MOD) * ((b + MOD) % MOD));
  }

  private static final Scanner scanner = new Scanner(System.in);
}