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)

def merge( tS, left, mid, right )
	n1 = mid - left;
	n2 = right - mid;
	tL = [];
	tR = [];
	for i in ( 0 ... n1 )
		tL.push( tS[ left+i ] )
	end
	for i in ( 0 ... n2 )
		tR.push( tS[ mid+i ] )
	end
	
	tL.push( 1000000001 )
	tR.push( 1000000001 )
	i = j = 0
	for k in ( left ... right )
		$cnt += 1
		if tL[i] <= tR[j]
			tS[k] = tL[i]
			i += 1
		else
			tS[k] = tR[j]
			j += 1
		end
	end
end

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

n = gets.to_i
arrS = gets.split( " " ).map { |val| val.to_i }
$cnt = 0
mergeSort( arrS, 0, n )
puts( arrS.join( " " ) )
puts( $cnt )