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 Ruby (judge verdict: Accepted)

def merge(a, left, mid, right)
    n1 = mid - left
    n2 = right - mid
    l = []
    r = []
    for i in 0..n1 - 1
        l[i] = a[left + i]
    end 
    for i in 0..n2 - 1
        r[i] = a[mid + i]
    end
    l[n1] = Float::INFINITY
    r[n2] = Float::INFINITY
    i = 0
    j = 0
    for k in left..right-1
        $n += 1
        if l[i] <= r[j] then
            a[k] = l[i]
            i += 1
        else
            a[k] = r[j]
            j += 1
        end
    end
    return a
end

def mergeSort(a, left, right)
    if left + 1 < right then
        mid = (left + right) / 2
        mergeSort(a, left, mid)
        mergeSort(a, mid, right)
        return merge(a, left, mid, right)
    else
        return a
    end
    
end

$n = 0
gets
s = gets.split.map(&:to_i)
left = 0
right = s.length
s = mergeSort(s, left, right)

puts s.join(" ")
puts $n