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 Scala (2.13.1) (judge verdict: Wrong Answer)

object Main extends App {
  val sc = new java.util.Scanner(System.in)
  val n, k = sc.nextInt()
  val a = Array.fill(n)(sc.nextLong())

  val mod = 1000000007

  def minus(a: Long, b: Long): Long = (a - b + mod) % mod
  def times(a: Long, b: Long): Long = (a * b) % mod

  val b = a.filter(_ != 0).map(e => (e.abs, e)).sorted
  if (b.length < k) {
    println(0)
  } else if (b.length == k) {
    if (b.count(_._2 < 0) % 2 == 0) {
      println(b.map(_._1).reduceLeft(times))
    } else {
      if (a.contains(0)) {
        println(0)
      } else {
        println(minus(0, b.map(_._1).reduceLeft(times)))
      }
    }
  } else {
    if (b.forall(_._2 > 0)) {
      println(b.takeRight(k).map(_._1).reduceLeft(times))
    } else if (b.forall(_._2 < 0)) {
      if (k % 2 == 0) {
        println(b.takeRight(k).map(_._1).reduceLeft(times))
      } else {
        if (a.contains(0)) {
          println(0)
        } else {
          println(minus(0, b.take(k).map(_._1).reduceLeft(times)))
        }
      }
    } else {
      val (r, c) = b.splitAt(b.length - k)
      if (c.count(_._2 < 0) % 2 == 0) {
        println(c.map(_._1).reduceLeft(times))
      } else {
        val pf = c.indexWhere(_._2 > 0)
        val mf = c.indexWhere(_._2 < 0)
        if (pf == -1) {
          c(mf) = r.findLast(_._2 > 0).get
        } else {
          val ml = r.findLast(_._2 < 0)
          val pl = r.findLast(_._2 > 0)
          var pd = Long.MaxValue
          var md = Long.MaxValue
          if (ml.isDefined) {
            pd = c(pf)._1 - ml.get._1
          }
          if (pl.isDefined) {
            md = c(mf)._1 - pl.get._1
          }
          if (pd < md) {
            c(pf) = ml.get
          } else {
            c(mf) = pl.get
          }
        }
        println(c.map(_._1).reduceLeft(times))
      }
    }
  }
}