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
#include <iostream>
#include <vector>
#include <algorithm>

int comp_count(const int &left, const int & right) {
	if (left + 1 < right) {
		return (right - left) + comp_count(left, (left + right) / 2) + comp_count((left + right) / 2, right);
	}
	else {
		return 0;
	}
}
int main() {
	int n;
	std::cin >> n;
	std::vector<int> vector(n);
	for (auto &v : vector) {
		std::cin >> v;
	}
	sort(vector.begin(), vector.end());
	std::cout << vector.at(0);
	for (auto i = 1; i < n; ++i) {
		std::cout << " " << vector.at(i);
	}
	std::cout << std::endl << comp_count(0, n) << std::endl;
	return 0;
}