#include <iostream>
#include <vector>
#include <cstdio>
using namespace std;
int data[1000010],n;
long long ans;
vector <int> G;
int main()
{
	cin >> n;
	for(int i = 1; i <= n; ++i)
		scanf("%d",&data[i]);
	for(int h = 1; ;)
	{
		if(h > n)
			break;
		G.push_back(h);
		h = h*3+1;
	}
	for(int i = G.size()-1; i >= 0; --i)
	{
		for(int j = G[i]+1; j <= n; ++j)
		{
			int v = data[j];
			int k = j-G[i];
			while((k > 0) && (data[k] > v))
			{
				data[k+G[i]] = data[k];
				k -= G[i];
				++ans;
			}
			data[k+G[i]] = v;
		}
	}
	cout << G.size() << endl << G[G.size()-1];
	for(int i = G.size()-2; i >= 0; --i)
		printf(" %d",G[i]);
	cout << endl << ans << endl;
	for(int i = 1; i <= n; ++i)
		printf("%d\n",data[i]);
	return 0;
}

### Prompt

Develop a solution in Cpp 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 <vector>
#include <cstdio>
using namespace std;
int data[1000010],n;
long long ans;
vector <int> G;
int main()
{
	cin >> n;
	for(int i = 1; i <= n; ++i)
		scanf("%d",&data[i]);
	for(int h = 1; ;)
	{
		if(h > n)
			break;
		G.push_back(h);
		h = h*3+1;
	}
	for(int i = G.size()-1; i >= 0; --i)
	{
		for(int j = G[i]+1; j <= n; ++j)
		{
			int v = data[j];
			int k = j-G[i];
			while((k > 0) && (data[k] > v))
			{
				data[k+G[i]] = data[k];
				k -= G[i];
				++ans;
			}
			data[k+G[i]] = v;
		}
	}
	cout << G.size() << endl << G[G.size()-1];
	for(int i = G.size()-2; i >= 0; --i)
		printf(" %d",G[i]);
	cout << endl << ans << endl;
	for(int i = 1; i <= n; ++i)
		printf("%d\n",data[i]);
	return 0;
}

```