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 <climits>
#include <iostream>
#include <iterator>
#include <queue>
#include <vector>
using namespace std;

int ans;

void mergeSort(vector<int>::iterator a, vector<int>::iterator b) {
	if (1 + a < b) {
		ans += b - a;
		mergeSort(a, next(a, (b - a) / 2));
		mergeSort(next(a, (b - a) / 2), b);
		queue<int> c, d;
		for (vector<int>::iterator i = a; i != next(a, (b - a) / 2); ++i) c.push(*i);
		c.push(INT_MAX);
		for (vector<int>::iterator i = next(a, (b - a) / 2); i != b; ++i) d.push(*i);
		d.push(INT_MAX);
		for (vector<int>::iterator i = a; i != b; ++i) {
			if (c.front() < d.front()) {
				*i = c.front();
				c.pop();
			}
			else {
				*i = d.front();
				d.pop();
			}
		}
	}
}

int main() {
	int n;
	cin >> n;
	vector<int> S(n);
	for (int& i : S) cin >> i;
	mergeSort(S.begin(), S.end());
	for (int i = 0; i < n - 1; ++i) cout << S[i] << ' ';
	cout << S.back() << endl << ans << endl;
}
