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(A, l, m, r):
    n1 = m - l
    n2 = r - m
    
    L = [A[l+i] for i in range(n1)]
    R = [A[m+i] for i in range(n2)]
    L.append(10**10)
    R.append(10**10)
    
    i = 0
    j = 0
    for k in range(l, r):
        if L[i]<=R[j]:
            A[k] = L[i]
            i += 1
        else:
            A[k] = R[j]
            j += 1
    global cnt
    cnt += i+j
    #print(cnt)
    return i+j
    
def mergeSort(A, l, r):
    if l+1<r:
        mid = (l + r)//2;
        mergeSort(A, l, mid)
        mergeSort(A, mid, r)
        merge(A, l, mid, r)


n = int(input())
A = list(map(int, input().split()))
#n = 10
#A = [8, 5, 9, 2, 6, 3, 7, 1, 10, 4]
cnt = 0

mergeSort(A, 0, n)
print(*A)
print(cnt)
