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_sort(a):
    return _merge_sort(a)

def _merge_sort(a):
    if len(a) == 1:
        return a,0
    m = len(a)/2
    b,count1 = _merge_sort(a[0:m])
    c,count2 = _merge_sort(a[m:len(a)])
    ret,count3 = merge(b,c)
    return (ret,count1+count2+count3)

def merge(b,c):
    a = []
    i = 0
    j = 0
    count = 0
    while i < len(b) and j < len(c):
        count += 1
        if b[i] > c[j]:
            a.append(c[j])
            j += 1
        else:
            a.append(b[i])
            i += 1
    if i == len(b):
        while j < len(c):
            count += 1
            a.append(c[j])
            j += 1
    elif j == len(c):
        while i < len(b):
            count += 1
            a.append(b[i])
            i += 1
    return a,count

n = int(raw_input())
a = map(int,raw_input().split(' '))
ret,count = merge_sort(a)
print ' '.join(map(str,ret))
print count