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)

$n = 0

def merge(sorted_left, sorted_right)
  arr = []
  l_p = r_p = 0
  max = 1000000000
  sorted_left.push(max)
  sorted_right.push(max)
  l_v = r_v = 0
  while l_v != max && r_v != max
    flag = true
    l_v = sorted_left[l_p].to_i
    while flag
      r_v = sorted_right[r_p].to_i
      if l_v == max && r_v == max then
        return arr
      elsif l_v < r_v then
        arr.push(l_v)
        l_p = l_p + 1
        flag = false
      else
        arr.push(r_v)
        r_p = r_p + 1
      end
      $n = $n + 1
    end
  end
  arr
end

def mergeSort(arr)
  if arr.size == 0 then
    arr
  elsif arr.size == 1 then
    arr
  else
    m = arr.size / 2
    left = arr.take(m)
    right = arr.drop(m)
    merge(mergeSort(left), mergeSort(right))
  end
end

n = STDIN.gets.to_i
arr = STDIN.gets.split(' ')
puts mergeSort(arr).join(' ')
puts $n