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 <bits/stdc++.h>
using namespace std;
int n;
int a[500005];
int b[500005];
int res=0;
void merge(int l,int r){
	int m=(l+r)/2;
	// l,m    m+1,r
	int p1=l,p2=m+1;
	for(int i=l;i<=r;i++){
		res++;
		if(p2>r||(p1<=m&&a[p1]<a[p2])){
			b[i]=a[p1];
			p1++;
		}else{
			b[i]=a[p2];
			p2++;
		}
	}
	for(int i=l;i<=r;i++)a[i]=b[i];
}
void ms(int l,int r){
	if(l>=r)return;
	ms(l,(l+r)/2);
	ms((l+r)/2+1,r);
	merge(l,r);
}
int main(){
	ios::sync_with_stdio(false);cin.tie(0);
	cin>>n;
	for(int i=1;i<=n;i++)cin>>a[i];
	ms(1,n);
	for(int i=1;i<n;i++)cout<<a[i]<<" ";
	cout<<a[n]<<endl<<res<<endl;
	return 0;
}
