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 <bits/stdc++.h>
using namespace std;

int n, ans = 0;
vector<int> a;

void solve(int l, int r);

int main() {
  cin >> n;
  a.resize(n);
  for(int i = 0; i < n; ++i) cin >> a[i];
  solve(0, n);
  for(int i = 0; i < n; ++i) {
    if(i != 0) cout << " ";
    cout << a[i];
  }
  cout << endl;
  cout << ans << endl;
  return 0;
}

void solve(int l, int r) {
  if(r - l <= 1) return;
  solve(l, (l + r) / 2);
  solve((l + r) / 2, r);
  vector<int> dummy(r - l, 0);
  int lid = l, rid = (l + r) / 2, half = (l + r) / 2;
  for(int i = 0; i < r - l; ++i) {
    ++ans;
    if(lid < half && rid < r) {
      if(a[lid] < a[rid])
        dummy[i] = a[lid++];
      else
        dummy[i] = a[rid++];
    }
    else if(lid < half)
      dummy[i] = a[lid++];
    else
      dummy[i] = a[rid++];
  }
  for(int i = 0; i < r - l; ++i) a[l + i] = dummy[i];
}
