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)


Notes

Constraints

* n ≤ 500000
* 0 ≤ an element in S ≤ 109

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.

Example

Input

10
8 5 9 2 6 3 7 1 10 4


Output

1 2 3 4 5 6 7 8 9 10
34
def merge( tS, left, mid, right ):
	global cnt
	n1 = mid - left;
	n2 = right - mid;
	tL = tS[ left : mid ];
	tR = tS[ mid : right ];	
	tL.append( 1000000001 )
	tR.append( 1000000001 )
	i = j = 0
	for k in range( left, right ):
		cnt += 1
		if tL[i] <= tR[j]:
			tS[k] = tL[i]
			i += 1
		else:
			tS[k] = tR[j]
			j += 1


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 )

n = int( raw_input() )
arrS = [ int( val ) for val in raw_input().split( ' ' ) ]
cnt = 0
mergeSort( arrS, 0, n )
print( " ".join( map( str, arrS ) ) )
print( cnt )