Problem:
Merge Sort

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)

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.

Constraints

n ≤ 500000

0 ≤ an element in S ≤ 109

Sample Input 1

10
8 5 9 2 6 3 7 1 10 4

Sample Output 1

1 2 3 4 5 6 7 8 9 10
34

Notes

Solution in Python3 (judge verdict: Accepted)

def merge(A, left, mid, right):
    global counter
    INFTY = 10 ** 10
    L = A[left:mid]
    R = A[mid:right]
    L.append(INFTY)
    R.append(INFTY)
    i, j = 0, 0
    for k in range(left, right):
        counter += 1
        if L[i] <= R[j]:
            A[k] = L[i]
            i += 1
        else:
            A[k] = R[j]
            j += 1

def merge_sort(A, left, right):
    if left + 1 < right:
        mid = (left + right) // 2
        merge_sort(A, left, mid)
        merge_sort(A, mid, right)
        merge(A, left, mid, right)
        
n = int(input())
A = list(map(int, input().split()))

counter = 0
merge_sort(A, 0, n)
print(' '.join(str(a) for a in A))
print(counter)