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, a[500005], l[500005], r[500005], inf = 1e9 + 1, s;
void merge( int x, int y ){
	if( x == y ) return;
	int c1 = (x + y) / 2;
	int c2 = ((x + y) / 2) + 1;
	
	merge( x, c1 );
	merge( c2 , y);
	
	for( int i = 0 ; i <= c1 - x ; i++ ) l[i] = a[i + x];
	for( int i = 0 ; i <= y - c2 ; i++ ) r[i] = a[i + c2];
	
	l[c1 - x + 1] = inf;
	r[y - c2 + 1] = inf;
	
	int k = x, l1 = 0, r1 = 0;
	
	for( int i = 0 ; i < y - x + 1 ; i++ ){
		a[k] = min( l[l1], r[r1]);
		k++;
		if( l[l1] > r[r1] ) r1++;
		else l1++;	
	}
	s += y - x + 1;	
	
}


int main(){
	
	cin >> n;
	for( int i = 0 ; i < n ; i++ ){
		cin >> a[i];
	}
	
	merge(0, n - 1);
	for( int i = 0 ; i < n ; i++ ){
		if( i != 0 ) cout << ' ';
		cout << a[i];
	}
	cout << "\n" << s << "\n";
	
}
