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<cstring>

using namespace std;

const int maxn=1e6;
int a[maxn];
int b[maxn];
int cnt;

void merge(int l,int mid,int r)
{
	int p1=l,p2=mid+1;//双指针
	int pos=l;//当前位置 
	while(p1<=mid&&p2<=r){
		cnt++;
		if(a[p1]<a[p2]){
			b[pos++]=a[p1++];
		}else{
			b[pos++]=a[p2++];
		}
	}
	while(p1<=mid){
		cnt++;
		b[pos++]=a[p1++];
	}
	while(p2<=r){
		cnt++;
		b[pos++]=a[p2++];
	}
	for(int i=l;i<=r;i++)a[i]=b[i];
}

void merge_sort(int l,int r)
{
	if(l<r){
		int mid=(l+r)/2;
		merge_sort(l,mid);
		merge_sort(mid+1,r);
		merge(l,mid,r);
	}
}

int main()
{
	memset(a,0,sizeof(a));
	memset(b,0,sizeof(b));
	int n;
	cin>>n;
	cnt=0;
	for(int i=1;i<=n;i++)cin>>a[i];
	merge_sort(1,n);
	for(int i=1;i<=n;i++){
		if(i>1)cout<<" ";
		cout<<a[i];
	}
	cout<<endl;
	cout<<cnt<<endl;
	return 0;
}


