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 cnt = 0;
int work[500010] = {0};

void mergeSort(int *a, int left, int right){
	if(left >= right) return;
	int mid = (right + left) / 2;
	mergeSort(a, left, mid);
	mergeSort(a, mid + 1, right);
	for(int i = left; i <= mid; i++)
		work[i] = a[i];
	for(int i = mid + 1; i <= right; i++)
		work[i] = a[right-(i-(mid+1))];
	int i = left, j = right;
	for(int k = left; k <= right; k++) {
		if(work[i] < work[j]) a[k] = work[i++];
		else a[k] = work[j--];
		cnt++;
	}
}

int main(){
	int n;
	scanf("%d", &n);
	int a[500010] = {0};
	for(int i = 0; i < n; i++)
		scanf("%d", &a[i]);
	mergeSort(a, 0, n-1);
	for(int i = 0; i < n - 1; i++)
		printf("%d ", a[i]);
	printf("%d\n", a[n-1]);
	printf("%d\n", cnt);
}

