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 OCaml (judge verdict: Accepted)

let cnt = ref 0

let merge a l m r =
  let ai = Array.make (m - l + 1) max_int in
  for i = 0 to m - l - 1 do
    ai.(i) <- a.(l + i)
  done;
  let aj = Array.make (r - m + 1) max_int in
  for i = 0 to r - m - 1 do
    aj.(i) <- a.(m + i)
  done;
  let i = ref 0 in
  let j = ref 0 in
  for k = l to r - 1 do
    incr cnt;
    if ai.(!i) <= aj.(!j) then begin
      a.(k) <- ai.(!i);
      incr i;
    end else begin
      a.(k) <- aj.(!j);
      incr j;
    end
  done

let rec merge_sort a l r =
  if (l + 1) < r then begin
    let m = (l + r) / 2 in
    merge_sort a l m;
    merge_sort a m r;
    merge a l m r
  end

let () =
  let n = read_int () in
  let a = Array.init n (fun _ -> Scanf.scanf "%d " (fun i -> i)) in
  merge_sort a 0 n;
  print_int a.(0);
  for i = 1 to n - 1 do
    print_string " ";
    print_int a.(i);
  done;
  print_newline ();
  print_int !cnt;
  print_newline ()