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>
using namespace std;

int cnt=0;
void ms(int *a, int left, int right, int *t) {
	if(right-left>1) {
		int m=left+(right-left)/2;
		int p=left, i=left, q=m;
		ms(a, left, m, t);
		ms(a, m, right, t);
		while(p<m || q<right) {
			cnt++;
			if(q>=right || (p<m && a[p]<a[q])) t[i++]=a[p++];
			else t[i++]=a[q++];
		}
		for(int i=left; i<right; i++) a[i]=t[i];
	}
}
const int LIM=1e6;
int a[LIM+10], tmp[LIM+10];
int main() {
	int n;
	cin>>n;
	for(int i=1; i<=n; i++) cin>>a[i];
	ms(a, 1, n+1, tmp);
	for(int i=1; i<=n; i++) {
		if(i>1) cout<<' ';
		cout<<a[i];
	}
	cout<<'\n'<<cnt<<'\n';
	return 0;
}
