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 <algorithm>

#pragma warning(disable: 4996)

#define MAX_N 500000

int v[MAX_N], dp[MAX_N + 7];

int main()
{
	register int N, C = 0;

	scanf("%d", &N);

	dp[N] = 1;

	for (int i = (N + 2) / 2; i > 1; i--)
	{
		dp[i] = dp[2 * i - 1] + 2 * dp[2 * i] + dp[2 * i + 1];
	}

	for (int i = 2; i <= N; i++)
	{
		C += dp[i] * i;
	}

	for (int i = 0; i < N; i++)
	{
		scanf("%d", &v[i]);
	}

	std::sort(v, v + N);

	for (int i = 0; i < N; i++)
	{
		if (i) { printf(" "); }

		printf("%d", v[i]);
	}

	printf("\n");

	printf("%d\n", C);

	return 0;
}