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 <cstdio>
#include <vector>
#include <algorithm>
using namespace std;

#define d(s, e) distance(s, e)
typedef vector<int> v;
typedef v::iterator vi;
int c = 0;

void mm(vi l, vi m, vi r)
{
	v L(d(l, m));
	copy(l, m, L.begin());
	vi cp = L.begin();

	for (vi i = l; i != r; i++) {
		if (cp == L.end()) {
			c += d(i, r);
			break;
		} else if (m == r) {
			c += d(i, r);
			copy(cp, L.end(), i);
			break;
		} else if(*cp <= *m)
			*i = *cp, cp++;
		else
			*i = *m, m++;
		c++;
	}


}

void ms(vi l, vi r)
{
	if (l + 1 < r) {
		vi m = l + d(l, r) / 2;
		ms(l, m);
		ms(m, r);
		mm(l, m, r);
	}
}

int main()
{
	int n;
	scanf("%d", &n);
	v S(n);
	for (int &x :S)
		scanf("%d", &x);

	ms(S.begin(), S.end());

	for (int &x: S)
		printf("%s%d", &x == &S[0] ? "": " ", x);
	puts("");
	printf("%d\n", c);
}
