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 <stdio.h>




int ans=0;

int a[500001];

int n;




void merge(int left,int mid,int right)

{

 int i,j,k;




 int n1=mid-left;

 int n2=right-mid;

 

 int l[n1+1],r[n2+1];




 for(i=0;i<n1;i++){

  l[i]=a[left+i];

 }

 for(i=0;i<n2;i++){

  r[i]=a[mid+i];

 }




 l[n1]=1000000001;

 r[n2]=1000000001;

 i=0;

 j=0;




 for(k=left;k<right;k++){

  ans++;

  if(l[i]<=r[j])

   a[k]=l[i++];

  else

   a[k]=r[j++];

 }

}




void mergeSort(int left,int right)

{

 int mid;




 if(left+1<right){

  mid=(left+right)/2;

  mergeSort(left,mid);

  mergeSort(mid,right);

  merge(left,mid,right);

 }

}




int main(void)

{

 int i;




 scanf("%d",&n);

 for(i=0;i<n;i++){

  scanf("%d",&a[i]);

 }

 

 mergeSort(0, n);




 for(i=0;i<n-1;i++){

  printf("%d ",a[i]);

 }

 printf("%d\n",a[n-1]);

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




 return 0;

}