#include <cstdio>
#include <vector>
#include <utility>
#include <algorithm>
#include <cmath>
using namespace std;

typedef vector<int> vi;

static inline void p(int x) { printf("%d\n", x); }

int main()
{

	int n, cnt = 0;
	scanf("%d", &n);

	vi A(n), G;
	for (int &x: A)
		scanf("%d", &x);

	for (int i = 1, x = 1; x <= n; x = (pow(3, ++i) - 1) / 2)
		G.push_back(x);
	reverse(G.begin(), G.end());

	for (int g: G)
		for (int i = g; i < n; i++) {
			int v = A[i], j = i - g;
			for(; j >= 0 && A[j] > v; j -= g)
				A[j+g] = A[j], cnt++;
			A[j+g] = v;
		}

	p(G.size());
	for (int &x: G)
		printf("%s%d", (&x == &G[0] ? "" : " "), x);
	puts("");
	p(cnt);
	for (int x: A)
		p(x);
}

### Prompt

Your task is to create a cpp solution to the following problem:

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 <cstdio>
#include <vector>
#include <utility>
#include <algorithm>
#include <cmath>
using namespace std;

typedef vector<int> vi;

static inline void p(int x) { printf("%d\n", x); }

int main()
{

	int n, cnt = 0;
	scanf("%d", &n);

	vi A(n), G;
	for (int &x: A)
		scanf("%d", &x);

	for (int i = 1, x = 1; x <= n; x = (pow(3, ++i) - 1) / 2)
		G.push_back(x);
	reverse(G.begin(), G.end());

	for (int g: G)
		for (int i = g; i < n; i++) {
			int v = A[i], j = i - g;
			for(; j >= 0 && A[j] > v; j -= g)
				A[j+g] = A[j], cnt++;
			A[j+g] = v;
		}

	p(G.size());
	for (int &x: G)
		printf("%s%d", (&x == &G[0] ? "" : " "), x);
	puts("");
	p(cnt);
	for (int x: A)
		p(x);
}

```