Problem:
Merge Sort

Write a program of a Merge Sort algorithm implemented by the following pseudocode. You should also report the number of comparisons in the Merge function.

Merge(A, left, mid, right)
n1 = mid - left;
n2 = right - mid;
create array L[0...n1], R[0...n2]
for i = 0 to n1-1
do L[i] = A[left + i]
for i = 0 to n2-1
do R[i] = A[mid + i]
L[n1] = SENTINEL
R[n2] = SENTINEL
i = 0;
j = 0;
for k = left to right-1
if L[i] <= R[j]
then A[k] = L[i]
i = i + 1
else A[k] = R[j]
j = j + 1

Merge-Sort(A, left, right){
if left+1 < right
then mid = (left + right)/2;
call Merge-Sort(A, left, mid)
call Merge-Sort(A, mid, right)
call Merge(A, left, mid, right)

Input

In the first line n is given. In the second line, n integers are given.

Output

In the first line, print the sequence S. Two consequtive elements should be separated by a space character.

In the second line, print the number of comparisons.

Constraints

n ≤ 500000

0 ≤ an element in S ≤ 109

Sample Input 1

10
8 5 9 2 6 3 7 1 10 4

Sample Output 1

1 2 3 4 5 6 7 8 9 10
34

Notes

Solution in Scala (judge verdict: Wrong Answer)

object Main {
  def main(args: Array[String]): Unit = {
    val (n, s) = (io.StdIn.readLine().toInt, io.StdIn.readLine().split(' ').map(_.toInt).toList)

    var cnt = 1
    def merge(left: List[Int], right: List[Int]): List[Int] = (left, right) match {
      case (x :: xs, y :: ys) if x <= y =>
        cnt += 1
        x :: merge(xs, right)
      case (x :: xs, y :: ys) if x > y =>
        cnt += 1
        y :: merge(left, ys)
      case (Nil, ys) =>
        cnt += 1
        right
      case (xs, Nil) =>
        cnt += 1
        left
    }

    def mergeSort(list: List[Int]): List[Int] = list match {
      case x :: Nil => list
      case _ =>
        val (l1, l2) = list.splitAt(list.size / 2)
        merge(mergeSort(l1), mergeSort(l2))
    }
    println(mergeSort(s).mkString(" "))
    println(cnt)
  }
}