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<cstring>
#include<algorithm>
using namespace std;
const int MAXN=500100;
int A[MAXN],T[MAXN];
int n;
int tot;
void input()
{
	cin>>n;
	for(int i=0;i<n;i++)
	    cin>>A[i];
}

void merge_sort(int l,int r)
{
	if(l==r) return;
	int mid=(l+r)/2;
	merge_sort(l,mid);
	merge_sort(mid+1,r);
	int x(l),y(mid+1),q(l);
	while(q<=r)
	{
		if((x<=mid && A[x]<=A[y]) || y>r) {T[q]=A[x];q++;x++;}
		else {T[q]=A[y];q++;y++;}
		tot++;
	}
	for(int i=l;i<=r;i++) A[i]=T[i];
}

void xxj()
{
	merge_sort(0,n-1);
}

void output()
{
	for(int i=0;i<n-1;i++)
	    cout<<A[i]<<' ';
    cout<<A[n-1]<<endl;
    cout<<tot<<endl;
}

int main()
{
	input();
	xxj();
	output();
	return 0;
}