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(input())
S=[int(i) for i in input().split()]
INF=10**10

def merge(A,l,m,r):
    cnt=0
    n1=m-l
    n2=r-m
    L,R=[0 for i in range(n1)],[0 for i in range(n2)]
    cnt=0
    L=A[l:l+n1]
    R=A[m:m+n2]
    L.append(INF)
    R.append(INF)
    i,j=0,0
    for k in range(l,r):
        cnt+=1
        if L[i]<=R[j]:
            A[k]=L[i]
            i+=1   
        else:
            A[k]=R[j]
            j+=1
    return cnt     

def mergesort(A,l,r):
    cnt=0
    if l+1<r:
        m=(l+r)//2
        cnt+=mergesort(A,l,m)
        cnt+=mergesort(A,m,r)
        cnt+=merge(A,l,m,r)
    return cnt


cnt=mergesort(S,0,n)
print(*S)
print(cnt)
