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>
using namespace std;
const int MAXN=5e5+7;
int s[MAXN],t[MAXN];
int cnt=0;
void merge (int l,int r,int mid) {
	int t1=l,t2=mid,pos=l;
	while (pos<r) {
		if (t2==r||(t1<mid&&s[t1]<s[t2])) t[pos++]=s[t1++];
		else t[pos++]=s[t2++];
		cnt++;
	}
	for (int i=l;i<r;i++) {
		s[i]=t[i];
	}
}
void mergeSort (int l,int r) {
	if (r-l<2) return;
	int mid=(l+r)>>1;
	mergeSort (l,mid);
	mergeSort (mid,r);
	merge (l,r,mid);
}
int main () {
	int n;
	scanf ("%d",&n);
	for (int i=0;i<n;i++) {
		scanf ("%d",&s[i]);
	}
	mergeSort (0,n);
	printf ("%d",s[0]);
	for (int i=1;i<n;i++) {
		printf (" %d",s[i]);
	}
	printf ("\n%d\n",cnt);
    return 0;
}
