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
global_count = 0

def sort(a):
    return merge_sort(a)

def merge_sort(a):
    n = len(a)
    if n <= 1:
        return a
    lo = 0
    hi = n
    med = (lo+hi)/2
    left = merge_sort(a[lo:med])
    right = merge_sort(a[med:hi])
    return merge(left, right)

def merge(a,b):
    global global_count
    c = []
    na = len(a)
    nb = len(b)
    i = 0
    j = 0
    while i < na and j < nb:
        if a[i] < b[j]:
            global_count += 1
            c.append(a[i])
            i += 1
        else:
            global_count += 1
            c.append(b[j])
            j += 1

    if i == na:
        while j < nb:
            global_count += 1
            c.append(b[j])
            j += 1
    if j == nb:
        while i < na:
            global_count += 1
            c.append(a[i])
            i += 1

    return c

n = int(raw_input())
a = map(int, raw_input().split(' '))
c = sort(a)
print ' '.join(map(str, c))
print global_count