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 merge(vector<int>& v) {
	int n = v.size();
	int h = n / 2;
	if (n == 1) return 0;
	vector<int> L(v.begin(), v.begin() + h);
	vector<int> R(v.begin() + h, v.end());
	int res = merge(L) + merge(R) + n;
	L.push_back(2e9);
	R.push_back(2e9);
	int l = 0, r = 0;
	for (int i = 0; i < n; i++) {
		if (L[l] <= R[r]) {
			v[i] = L[l++];
		} else {
			v[i] = R[r++];
		}
	}
	return res;
}

int main() {
	int n;
	cin >> n;
	vector<int> a(n);
	for (auto& x:a) cin >> x;
	int r = merge(a);
	for (int i = 0; i < n; i++) {
		if (i) cout << " ";
		cout << a[i];
	}
	cout << endl << r << endl;
}
