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: WA: Presentation Error)

def merge(a,n)
  x=0
  left=Array.new
  right=Array.new 
  l=(n/2).to_i
  r=n-l
  l.times do |i|
    left<<a[i]
  end
  r.times do |i|
    right<<a[i+l]
  end
  left<<10**9+1
  right<<10**9+1
  if l>1 then
    x+=merge(left,l)
  end
  if r>1 then
    x+=merge(right,r)
  end
  i=0
  j=0
  n.times do |k|
    if left[i]<right[j] then
      a[k]=left[i]
      i+=1
    else
      a[k]=right[j]
      j+=1
    end
  end
  return i+j+x
end

n=gets.to_i
a=gets.split.map!{|i|i.to_i}
a.slice!(n..-1)
x=merge(a,n)
n.times do |i|
  print a[i]," "
end
print "\n",x