#include<stdio.h>

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

  return cnt;
}

void shellSort(int A[], int n) {
  int im, m = 1, v = 1, G[100], sum = 0;
  for ( im = 0; v <= n; im++ ) {
    G[im] = v;
    m *= 3;
    v += m;
  }

  printf("%d\n", im);
  for ( int i = im - 1; i >= 0; i-- ) {
    if ( i != (im - 1) ) printf(" ");
    printf("%d", G[i]);
    sum += insertionSort(A, n, G[i]);
  }

  printf("\n%d\n", sum);
  for ( int i = 0; i < n; i++ ) {
    printf("%d\n", A[i]);
  }
}

int main() {
  int n, A[1000000];
  scanf("%d", &n);

  for ( int i = 0; i < n; i++ ) scanf("%d", &A[i]);

  shellSort(A, n);

  return 0;
}


### Prompt

Create a solution in CPP for 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<stdio.h>

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

  return cnt;
}

void shellSort(int A[], int n) {
  int im, m = 1, v = 1, G[100], sum = 0;
  for ( im = 0; v <= n; im++ ) {
    G[im] = v;
    m *= 3;
    v += m;
  }

  printf("%d\n", im);
  for ( int i = im - 1; i >= 0; i-- ) {
    if ( i != (im - 1) ) printf(" ");
    printf("%d", G[i]);
    sum += insertionSort(A, n, G[i]);
  }

  printf("\n%d\n", sum);
  for ( int i = 0; i < n; i++ ) {
    printf("%d\n", A[i]);
  }
}

int main() {
  int n, A[1000000];
  scanf("%d", &n);

  for ( int i = 0; i < n; i++ ) scanf("%d", &A[i]);

  shellSort(A, n);

  return 0;
}


```