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[500010], b[500010], cnt = 0;
void merge(int l, int mid, int r){
	int n1 = mid - l, n2 = r - mid;
	int T = r - l, i = 0, j = 0, now = 0;
	while(T--){
		cnt++;
		if(i == n1){
			b[now++] = a[mid + j];
			j++;
		}
		else if(j == n2){
			b[now++] = a[l + i];
			i++;
		}
		else if(a[l + i] < a[mid + j]){
			b[now++] = a[l + i];
			i++;
		}
		else{
			b[now++] = a[mid + j];
			j++;
		}
		
	}
	for(int i = 0; i < now; i++) a[l + i] = b[i];
	//for(int i = 0; i < now; i++) printf("%d ", b[i]);
	//printf("\n");
}
void MergeSort(int l, int r){
	if(l + 1 >= r) return;
	int mid = (l + r) / 2;
	MergeSort(l, mid);
	MergeSort(mid, r);
	merge(l, mid, r);
}
int main()
{
	scanf("%d", &n);
	for(int i = 0; i < n; i++) scanf("%d", &a[i]);
	MergeSort(0, n);
	for(int i = 0; i < n - 1; i++) printf("%d ", a[i]);
	
	printf("%d\n%d\n", a[n-1], cnt);
	return 0; 
}
