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: Accepted)

#
# merge sort
#   [left, right)
#

$comperation_count = 0

def merge_sort(arr)
  return arr if arr.size <= 1
  mid = arr.size / 2
  left = merge_sort(arr[0...mid])
  right = merge_sort(arr[mid...arr.size])
  ret = []
  while !left.empty? || !right.empty?
    $comperation_count += 1
    if left.empty?
      ret << right.shift
    else
      if right.empty?
        ret << left.shift
      else
        if left[0] <= right[0]
          ret << left.shift
        else
          ret << right.shift
        end
      end
    end
  end
  ret
end


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

puts merge_sort(s) * " "
puts $comperation_count