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
n = int(raw_input())
A = map(int, raw_input().split(" "))
count = 0

def merge(left, right):
    global count
    
    sorted_list = list()
    left_index = 0
    right_index = 0

    while left_index < len(left) and right_index < len(right):
        if left[left_index] <= right[right_index]:
            sorted_list.append(left[left_index])
            left_index += 1

        else:
            sorted_list.append(right[right_index])
            right_index += 1
            
        count += 1

    if left:
        sorted_list.extend(left[left_index:])
        count += len(left[left_index:])
    if right:
        sorted_list.extend(right[right_index:])
        count += len(right[right_index:])
        

    return sorted_list


    
def mergeSort(A):
    if len(A) <= 1:
        return A

    mid = len(A) // 2
    left = A[:mid]
    right = A[mid:]

    left = mergeSort(left)
    right = mergeSort(right)

    return list(merge(left, right))

print " ".join(map(str, mergeSort(A)))
print count