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)

INFTY = 9999999999
$cnt = 0

def merge(a, left, mid, right)
	n1 = mid - left
	n2 = right - mid
	l = []
	r = []
	0.upto(n1-1) do |i|
		l[i] = a[left + i]
	end
	0.upto(n2-1) do |i|
		r[i] = a[mid + i]
	end
	l[n1] = INFTY
	r[n2] = INFTY
	i = 0
	j = 0
	left.upto(right-1) do |k|
		$cnt += 1
		if l[i] <= r[j]
			a[k] = l[i]
			i += 1
		else
			a[k] = r[j]
			j += 1
		end
	end
end

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

n = gets.to_i
s = gets.split.map &:to_i
mergeSort(s, 0, s.size)
puts s.join(" ")
puts $cnt