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 cnt=0;
void merge_sort(vector<int> &a) {
    int n = a.size();
    if (n==1) return;

    vector<int> b(begin(a), begin(a) + n/2);
    vector<int> c(begin(a) + n/2, end(a));
    merge_sort(b);
    merge_sort(c);

    for (int i = 0, j = 0, k = 0; i < n; ++i) {
        if (k == c.size())      a[i] = b[j++];
        else if (j == b.size()) a[i] = c[k++];
        else if (b[j] < c[k])   a[i] = b[j++];
        else                    a[i] = c[k++];
        cnt++;
    }
}
signed main() {
    int n;
    cin >> n;
    vector<int> a(n);
    for (int i = 0; i < n; ++i) cin >> a[i];
    merge_sort(a);
    for (int i = 0; i < n; ++i) {
        cout << a[i];
        cout << (i != n -1 ? " " : "\n");
    }
    cout << cnt << endl;
    return 0;
}