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>
#include <vector>

using namespace std;

int cnt;

void Merge(vector<int> &A){
  int n = (int)A.size();
  
  if(n > 1){
    vector<int> L(A.begin(), A.begin()+n/2);
    vector<int> R(A.begin()+n/2, A.end());

    Merge(L);
    Merge(R);

    int l = (int)L.size(), r = (int)R.size();
    for(int i = 0,j = 0,k = 0 ; i < n ; i++){
      cnt++;
      if(k == r) A[i] = L[j++];
      else if(j == l) A[i] = R[k++]; 
      else if(L[j] < R[k]) A[i] = L[j++];
      else A[i] = R[k++];
    }
  }
  
}

int main(){
  int n;
  
  cin >> n;
  vector<int> v(n);
  for(int i = 0 ; i < n ; i++){
    cin >> v[i];
  }

  Merge(v);

  int len = (int)v.size();
  for(int i = 0 ; i < len ; i++){
    if(i == len-1) cout << v[i] << endl;
    else cout << v[i] << " ";
  }
  cout << cnt << endl;

  return 0;
}