#include<iostream>
#include<stdio.h>
#include<math.h>
#include<string.h>
#include<cmath>
#include<vector>
#include<cstdio>
#include<algorithm>
using namespace std;
long long cnt;
int l;
int a[1000000];
int n;
vector<int>g;
void insertionSort(int a[],int n,int g){
for(int i=g;i<n;i++){
int v = a[i];
int j = i - g ;
while (j >= 0 && a[j] > v){
a[j+g] = a[j];
j = j - g;
cnt++;
}
a[j+g] = v;
}
}

void shellSort(int a[],int n){
for(int h=1; ;){
if(h>n) break;
g.push_back(h);
h=3*h+1;
}
for(int i=g.size()-1;i>=0;i--){
insertionSort(a,n,g[i]);
}
}
int main(){
cin>>n;
for(int i=0;i<n;i++) scanf("%d",&a[i]);
cnt=0;
shellSort(a,n);
cout<<g.size()<<endl;
for(int i=g.size()-1;i>=0;i--){
    printf("%d",g[i]);
    if(i) printf(" ");
}
printf("\n");
printf("%d\n",cnt);
for(int i=0;i<n;i++) printf("%d\n",a[i]);
return 0;
}
### Prompt

Please provide a Cpp coded solution to the problem described below:

Shell Sort

Shell Sort is a generalization of Insertion Sort to arrange a list of $n$ elements $A$.


1  insertionSort(A, n, g)
2      for i = g to n-1
3          v = A[i]
4          j = i - g
5          while j >= 0 && A[j] > v
6              A[j+g] = A[j]
7              j = j - g
8              cnt++
9          A[j+g] = v
10
11 shellSort(A, n)
12     cnt = 0
13     m = ?
14     G[] = {?, ?,..., ?}
15     for i = 0 to m-1
16         insertionSort(A, n, G[i])


A function shellSort(A, n) performs a function insertionSort(A, n, g), which considers every $g$-th elements. Beginning with large values of $g$, it repeats the insertion sort with smaller $g$.

Your task is to complete the above program by filling ?. Write a program which reads an integer $n$ and a sequence $A$, and prints $m$, $G_i (i = 0, 1, ..., m − 1)$ in the pseudo code and the sequence $A$ in ascending order. The output of your program must meet the following requirements:

* $1 \leq m \leq 100$
* $0 \leq G_i \leq n$
* cnt does not exceed $\lceil n^{1.5}\rceil$

Constraints

* $1 \leq n \leq 1,000,000$
* $0 \leq A_i \leq 10^9$

Input

In the first line, an integer $n$ is given. In the following $n$ lines, $A_i (i=0,1,...,n-1)$ are given for each line.

Output

In the first line, print an integer $m$. In the second line, print $m$ integers $G_i (i=0,1,...,m-1)$ separated by single space character in a line.
In the third line, print cnt in a line. In the following $n$ lines, print $A_i (i=0,1,...,n-1)$ respectively.

This problem has multiple solutions and the judge will be performed by a special validator.

Examples

Input

5
5
1
4
3
2


Output

2
4 1
3
1
2
3
4
5


Input

3
3
2
1


Output

1
1
3
1
2
3

### Response

```cpp
#include<iostream>
#include<stdio.h>
#include<math.h>
#include<string.h>
#include<cmath>
#include<vector>
#include<cstdio>
#include<algorithm>
using namespace std;
long long cnt;
int l;
int a[1000000];
int n;
vector<int>g;
void insertionSort(int a[],int n,int g){
for(int i=g;i<n;i++){
int v = a[i];
int j = i - g ;
while (j >= 0 && a[j] > v){
a[j+g] = a[j];
j = j - g;
cnt++;
}
a[j+g] = v;
}
}

void shellSort(int a[],int n){
for(int h=1; ;){
if(h>n) break;
g.push_back(h);
h=3*h+1;
}
for(int i=g.size()-1;i>=0;i--){
insertionSort(a,n,g[i]);
}
}
int main(){
cin>>n;
for(int i=0;i<n;i++) scanf("%d",&a[i]);
cnt=0;
shellSort(a,n);
cout<<g.size()<<endl;
for(int i=g.size()-1;i>=0;i--){
    printf("%d",g[i]);
    if(i) printf(" ");
}
printf("\n");
printf("%d\n",cnt);
for(int i=0;i<n;i++) printf("%d\n",a[i]);
return 0;
}
```