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 a[500005];
int tmp[500005];
int  n, cnt;
void Solve(int a[],int l ,int mid ,int r , int tmp[]) {
	//合并，使用三个指针指向三个数组
	int p = 0;
	int p1 = l, p2 = mid + 1;
	while (p1 <= mid && p2 <= r) {
		cnt++;
		if (a[p1] <= a[p2])
			tmp[p++] = a[p1++]; 
		else
			tmp[p++] = a[p2++];
	}
	while (p1 <= mid) {
		cnt++;
		tmp[p++] = a[p1++];
	}
	while (p2 <= r) {
		cnt++;
		tmp[p++] = a[p2++];
	}
	for (int i = 0; i <r-l+1; i++) {
		a[l+i] = tmp[i];
	}
}
void MerageSort(int a[] ,int l ,int r ,int tmp[]) {
	if (l < r) {
		int mid = l + (r - l) / 2;
		MerageSort(a, l, mid, tmp);
		MerageSort(a, mid + 1, r, tmp);
		Solve(a, l, mid, r, tmp);//合并函数
	}
}
int main() {
	cin >> n;
	for (int i = 0; i < n; i++)
		cin >> a[i];
	MerageSort(a, 0, n - 1, tmp);
	for (int i = 0; i < n; i++) {
		if (i) cout << " ";
		cout << a[i];
	}
	cout << endl << cnt << endl;
}
