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<cstdio>
#include<algorithm>
using namespace std;
int a[500000];
int all = 0;
int b[250000], c[250000];
void merge_sort(int left, int midle, int right) {
	for (int i = 0; i < midle - left; i++) {
		b[i] = a[i + left];
	}
	for (int i = 0; i < right - midle; i++) {
		c[i] = a[i + midle];
	}
	int j = 0;
	int k = 0;
	for (int i = left; i < right; i++) {
		if (k==right-midle||(j<midle-left&& b[j]<=c[k]))
		{
			a[i] = b[j];
			j++;
		}
		else{
			a[i] = c[k];
			k++;
		}
		all++;
	}
}
void merge(int left, int right) {
	if (left+1 <right)
	{
		int mid = (left + right) / 2;
		merge(left, mid);
		merge(mid, right);
		merge_sort(left, mid, right);
	}
}
int main(void) {
	int n;
	cin >> n;
	for (int i = 0; i < n; i++) {
		scanf("%d", &a[i]);
	}
	merge(0, n);
	for (int i = 0; i < n; i++) {
		if (i) cout << " ";
		cout << a[i];
	}
	cout << endl << all << endl;
}
