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 INFTY 1000000001

using namespace std;
int A[500001], L[500001], R[500001];
int n;
int countNum = 0;

void merge(int left, int mid, int right){
	int n1 = mid - left;
	int n2 = right - mid;
	for(int i = 0; i < n1; i++){
		L[i] = A[left + i];
	}
	for(int i = 0; i < n2; i++){
		R[i] = A[mid + i];
	}
	L[n1] = INFTY;
	R[n2] = INFTY;
	int i = 0;
	int j = 0;
	for(int k = left; k < right; k++){
		countNum++;
		if(L[i] <= R[j]){
			A[k] = L[i];
			i++;
		}else{
			A[k] = R[j];
			j++;
		}
	}
}

void mergeSort(int left , int right){
	if(left + 1 < right){
		int mid = (left + right) / 2;
		mergeSort(left, mid);
		mergeSort(mid, right);
		merge(left, mid, right);
	}
}


int main(){
	cin >> n;
	for(int i = 0; i < n; i++){
		cin >> A[i];
	}
	mergeSort(0, n);
	for(int i = 0; i < n - 1; i++){
		cout << A[i] <<" ";
	}
	cout<<A[n - 1] << endl;
	cout << countNum << endl;
	return 0;
}