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 = gets.to_i
S = gets.split.map(&:to_i)

$inf = 1.0 / 0.0
$count = 0

def merge_sort(arr)
  left = arr[0..(arr.length / 2 - 1)]
  right = arr[(arr.length / 2)..(arr.length - 1)]
  left = merge_sort(left) unless left.length <= 1
  right = merge_sort(right) unless right.length <= 1
  left.push($inf)
  right.push($inf)
  res = []
  l, r = 0, 0
  while !(l == left.length || r == right.length)
    $count += 1
    if left[l] < right[r]
      res.push(left[l])
      l += 1
    else
      res.push(right[r])
      r += 1
    end
  end
  $count -= 1
  res.pop
  return res
end

puts merge_sort(S).join(' ')
puts $count