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)

INF = 10**9 + 1

def merge(a,left,mid,right)
  count = 0
  lSize = mid-left;
  rSize = right-mid;
  l = []
  r = []
  (0...lSize).each do |i|
    l[i] = a[left+i]
  end
  (0...rSize).each do |i|
    r[i] = a[mid+i]
  end
  l[lSize] = INF
  r[rSize] = INF
  i,j=0,0
  (left...right).each do |k|
    count += 1
    if l[i] <= r[j]
      a[k] = l[i]
      i += 1
    else 
      a[k] = r[j]
      j += 1
    end
  end
  return count
end

def mergeSort(a,left,right)
  count = 0
  if left+1 < right
    mid = (left+right)/2;
    count += mergeSort(a,left,mid)
    count += mergeSort(a,mid,right)
    count += merge(a,left,mid,right)
  end
  return count
end

n = gets.to_i
a = gets.split.map(&:to_i)
count = mergeSort(a,0,n)
printf("%s\n%d\n",a.join(" "),count)