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 = list(map(int,input().split()))

def marge(l,r):
    count = 0
    conq = []
    i = 0
    j = 0
    k = len(l) + len(r)
    l.append(float('inf'))
    r.append(float('inf'))
    for o in range(k):
        if l[i] <= r[j]:
            conq.append(l[i])
            i += 1
        else:
            conq.append(r[j])
            j += 1
        count += 1
    return conq,count
           
def margeSort(s):
    n = len(s)
    if n == 1:
        return s,0
    else:
        L, cnt_L = margeSort(s[:n//2])
        R,cnt_R = margeSort(s[n//2:])
        s,cnt = marge(L,R)         
    return s,cnt+cnt_R+cnt_L
S,count = margeSort(S)
print(*S)
print(count)
