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
cnt = 0
def merge(A,left,mid,right):
	'''n1 = mid - left
	n2 = right - mid
	L = []
	R = []
	for i in range(0,n1):
		L.append(A[left+i])

	for i in range(0,n2):
		R.append(A[mid+i])
	L.append(1000000001)
	R.append(1000000001)
	'''
	L = A[left:mid]+[1000000001]
	R = A[mid:right]+[1000000001]

	i = 0
	j = 0
	global cnt
	for k in range(left,right):
		cnt += 1
		if L[i] <= R[j]:
			A[k]=L[i]
			i += 1
		else:
			A[k] = R[j]
			j += 1

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)


if __name__ == '__main__':
	n = (int)(input())
	a = list(map(int,input().split()))
	
	mergeSort(a,0,n)
	print(*a)
	print(cnt)