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>
# define N 10000000
# define INF 1000000000
using namespace std;

int n, S[N], cnt=0;
void merge(int left, int mid, int right){
  int n1 = mid - left;
  int n2 = right - mid;
  int L[n1], R[n2];

  for(int i=0; i<n1; i++) L[i] = S[left+i];
  for(int i=0; i<n2; i++) R[i] = S[mid+i];
  L[n1] = INF;
  R[n2] = INF;
  
  for(int k=left, i=0, j=0; k<right; k++){
    if(L[i] <= R[j]){
      S[k] = L[i];
      i++;
    }else{
      S[k] = R[j];
      j++;
    }
    cnt++;
  }
  return;
}
void merge_sort(int left, int right){
  if(left+1 < right){
    int mid = (left + right)/2;
    merge_sort(left, mid);
    merge_sort(mid, right);
    merge(left, mid, right);
  }
  return;
}
int main(){
    cin >> n;
    for(int i=0; i<n; i++) scanf("%d",&S[i]);
    merge_sort(0, n);
    for(int i=0; i<n-1; i++) printf("%d ",S[i]);
    printf("%d\n",S[n-1]);
    printf("%d\n",cnt);
    return 0;
}

