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 count
    n1 = mid - left
    n2 = right - mid
    L = a[left:left + n1] + [SENTINEL]
    R = a[mid:mid + n2] + [SENTINEL]
    i = j = 0
    for k in range(left, right):
        if L[i] <= R[j]:
            a[k] = L[i]
            i += 1
        else:
            a[k] = R[j]
            j += 1
    count += right - left

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)

SENTINEL = 1000000001
n = int(input())
a = list(map(int, input().strip().split()))
count = 0
merge_sort(a, 0, len(a))
print(' '.join(map(str, a)))
print(count)