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)

n = STDIN.gets.to_i
input = STDIN.gets.split.map(&:to_i)
@count = 0
def merge_sort(array)
  return array if array.size == 1
  mid_index = array.size/2
  left = array[0..mid_index-1]
  right = array[mid_index..-1]
  sorted_left = merge_sort(left)
  sorted_right = merge_sort(right)
  merge(sorted_left, sorted_right)
end
def merge(left, right)
  result = []
  until left.empty? && right.empty? do
    @count += 1
   if left.empty?
     result << right.shift
   elsif right.empty?
     result << left.shift
   elsif left[0] < right[0]
     result << left.shift
   else
     result << right.shift
   end
  end
  result
end
puts merge_sort(input)*" "
p @count