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<iostream>
using namespace std;
string s,s1,s2;
int t,n,m,tmp,cnt=0;
int a[500001];
int x[500001],y[500001];
void merge_sort(int l,int r){
	 int mid,n1,n2;
	 if (r-l==1) return;
	 mid=(l+r)/2; 
	 merge_sort(l,mid);
	 merge_sort(mid,r);
	 n1=mid-l; n2=r-mid;
	 for (int i=0; i<n1; i++)
	     x[i]=a[l+i];
	 for (int i=0; i<n2; i++)
	     y[i]=a[mid+i];
	 x[n1]=2000000001; y[n2]=2000000001;
	 int i=0,j=0;
	 for (int k=l; k<r; k++){
	 	 if (x[i]<y[j])
	 	 	a[k]=x[i++];
	 	 else{
	 	 	a[k]=y[j++];
		  }
	 	  cnt++;
	 }
   
}
int main(){
	ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);
    cin>>n;
    for (int i=0; i<n; i++)
        cin>>a[i];
    merge_sort(0,n);
    cout<<a[0];
    for (int i=1; i<n; i++)
        cout<<' '<<a[i];
    cout<<'\n';
    cout<<cnt<<endl;
	return 0;
} 
