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>
#define rep(i,n) for(int i=0;i<n;i++)
using namespace std;
using ll = long long;
static const int INF=2e9;
int cnt=0;

void merge(int A[],int left,int mid,int right){
    int n1=mid-left,n2=right-mid;
    int B[n1+1],C[n2+1];
    B[n1]=C[n2]=INF;
    rep(i,n1)B[i]=A[left+i];
    rep(i,n2)C[i]=A[mid+i];
    int j=0,k=0;
    for(int i=left;i<right;i++){
        cnt++;
        if(B[j]>C[k])A[i]=C[k++];
        else A[i]=B[j++];
    }
    return;
}

void mergesort(int A[],int left,int right){
    if(left+1>=right)return;
    int mid=(left+right)/2;
    mergesort(A,left,mid);
    mergesort(A,mid,right);
    merge(A,left,mid,right);
    return;
}

int main(){
    int n;cin>>n;
    int A[n];
    rep(i,n)scanf("%d",&A[i]);
    mergesort(A,0,n);
    rep(i,n){
        if(i!=0)cout<<" ";
        cout<<A[i];
    }
    cout<<"\n"<<cnt<<"\n";
    return 0;
}
