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>
#define rep(i,o,u) for(int i = o;i < u; i++)
using namespace std;
int L[500000/2+2],R[500000/2+2];
int ans = 0;
void merge(int a[],int left,int mid,int right){
	int n1 = mid - left;
	int n2 = right - mid;
	rep(i,0,n1) L[i] = a[left + i];
	rep(i,0,n2) R[i] = a[mid + i];
	L[n1] = R[n2] = 2000000000;
	int i = 0,j = 0;
	rep(k,left,right){
		ans++;
		if(L[i] <= R[j]){
			a[k] = L[i++];
		}else{
			a[k] = R[j++];
		}
	}
}
void mergeSort(int a[],int left,int right){
	if(left + 1 < right){
		int mid = (left + right) / 2;
		mergeSort(a,left,mid);
		mergeSort(a,mid,right);
		merge(a,left,mid,right);
	}
}
int main(){
	int n;
	cin >> n;
	int a[500001];
	rep(i,0,n) cin >> a[i];
	mergeSort(a,0,n);
	rep(i,0,n-1){cout << a[i] << " ";}
	cout << a[n-1] << endl;
	cout << ans << endl;
	return 0;
}