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>

int n,a[500000],sw=0;


void merge(int left,int mid,int right){
    int l[500000],r[500000],i,j,k,l1,r1;
    j=k=0;
    l1=mid-left;
    r1=right-mid;

    for(i=0;i<l1;i++)l[i]=a[left+i];
    for(i=0;i<r1;i++)r[i]=a[mid+i];

    l[l1]=r[r1]=1000000000;

    for(i=left;i<right;i++){
        sw++;
        if(l[j]>=r[k]){
            a[i]=r[k];
            k++;
        }
        else {
            a[i]=l[j];
            j++;
        }
    }
}


void mergeSort(int left,int right){
    int mid;
    if(left+1<right){
    mid=(left+right)/2;
    mergeSort(left,mid);
    mergeSort(mid,right);
    merge(left,mid,right);
    }
}


int main(){
    int i;

    scanf("%d",&n);

    for(i=0;i<n;i++){
        scanf("%d",&a[i]);
    }

    mergeSort(0,n);

    for(i=0;i<n;i++){
        if(i)printf(" ");
        printf("%d",a[i]);
    }
    printf("\n%d\n",sw);
}
