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<stdio.h>
#include<iostream>
#include<algorithm>
#include<stdlib.h>

using namespace std;

const int maxn=500100;

int t[maxn],a[maxn];

int c=0;

void merge_sort(int *a,int x,int y,int *t){
	if(y-x>1){
		int m=(x+y)/2;
		int p=x,q=m,i=x;
		merge_sort(a,x,m,t);
		merge_sort(a,m,y,t);
		while(p<m || q<y){
			if(q>=y || (p<m && a[p]<=a[q]))
			{
				c++;
				t[i++]=a[p++];
			}
			else
			{
				c++;
				t[i++]=a[q++];
			}
		}
	for(i = x;i<y;i++)
		a[i]=t[i];
	}
}

int main()
{
	int n;
	while(scanf("%d",&n)!=EOF){
		for(int i=0;i<n;i++)
		{
			scanf("%d",&a[i]);
		}
		int l=0,r=n;
		merge_sort(a,l,r,t);
		for(int i=0;i<n;i++)
		{
			if(i!=0)
				printf(" ");
			printf("%d",a[i]);
		}
		printf("\n");
		printf("%d\n",c);
	}
	return 0;
}

