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(left_array, right_array)
  return_array = []
  left_index, right_index = 0, 0
  left_array << 10**9
  right_array << 10**9
  0.upto(left_array.length + right_array.length - 3) do
    if left_array[left_index] < right_array[right_index]
      return_array << left_array[left_index]
      left_index += 1
    else
      return_array << right_array[right_index]
      right_index += 1
    end
  end
  return_array
end

def merge_sort(target_array)
  right = target_array.length
  mid = target_array.length / 2
  left_array = target_array[0...mid]
  right_array = target_array[mid...right]
  return target_array if target_array.length == 1
  return merge(merge_sort(left_array), merge_sort(right_array))
end

gets.to_i
input = gets.split(' ').map(&:to_i)
puts merge_sort(input).join ' '