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 Ruby (judge verdict: Wrong Answer)

def merge_sort(arr, &block)
  if arr.size <= 1
    block.call
    return arr
  end

  mid = arr.size / 2

  arr1 = arr[0...mid]
  arr2 = arr[mid..-1]

  merge(merge_sort(arr1, &block), merge_sort(arr2, &block), &block)
end

def merge(arr1, arr2, &block)
  ret = []

  until arr1.empty? || arr2.empty?
    if arr1.first <= arr2.first
      ret << arr1.shift
    else
      ret << arr2.shift
    end

    block.call
  end

  ret.push(*arr1) unless arr1.empty?
  ret.push(*arr2) unless arr2.empty?

  ret
end

n = gets.to_i
arr = gets.split.map(&:to_i)

count = 0

puts merge_sort(arr) { count += 1 }.join(' ')

puts count