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)

def merge as, left, mid, right
  l_as = as.slice(left..mid)
  r_as = as.slice((mid + 1)..right)
  l_as << Float::INFINITY
  r_as << Float::INFINITY

  l_i, r_i = [0, 0]
  cnt = 0
  left.upto(right) do |i|
    cnt += 1
    if l_as[l_i] <= r_as[r_i]
      as[i] = l_as[l_i]
      l_i += 1
    else
      as[i] = r_as[r_i]
      r_i += 1
    end
  end
  return cnt
end

def merge_sort as, left, right
  cnt = 0
  if left < right
    mid = (left + right) / 2
    cnt += merge_sort as, left, mid
    cnt += merge_sort as, mid + 1, right
    cnt += merge as, left, mid, right
  end
  return cnt
end

n = gets.chomp.to_i
as = gets.chomp.split(' ').map(&:to_i)
cnt = merge_sort as, 0, n - 1
puts as.join(' ')
puts cnt