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<stdio.h>
using namespace std;

int a[500005],l[250005],r[250005],x=0;

void s(int a[],int n,int left,int mid,int right){
    int n1=right-mid;
    int n2=mid-left;
    for(int i=0;i<n1;i++) r[i]=a[i+mid];
    for(int i=0;i<n2;i++) l[i]=a[i+left];
    r[n1]=l[n2]=2000000000;
    int i=0,j=0;
    for(int g=left;g<right;g++){
        x++;
        if(l[i]<=r[j]){
            a[g]=l[i++];
        }
        else{
            a[g]=r[j++];
        }
    }
}

void sm(int a[],int n,int left,int right){
    if(left+1<right){
        int mid=(right+left)/2;
        sm(a,n,left,mid);
        sm(a,n,mid,right);
        s(a,n,left,mid,right);
    }
}

int main(){
    int n;
    scanf("%d",&n);
    for(int i=0;i<n;i++) scanf("%d",&a[i]);
    sm(a,n,0,n);
    for(int i=0;i<n;i++){
        if(!i) printf("%d",a[i]);
        else printf(" %d",a[i]);
    }
    printf("\n%d\n",x);
    return 0;
}