Problem: You are given a deck of n cards numbered from 1 to n (not necessarily in this order in the deck). You have to sort the deck by repeating the following operation. 

  * Choose 2 ≤ k ≤ n and split the deck in k nonempty contiguous parts D_1, D_2,..., D_k (D_1 contains the first |D_1| cards of the deck, D_2 contains the following |D_2| cards and so on). Then reverse the order of the parts, transforming the deck into D_k, D_{k-1}, ..., D_2, D_1 (so, the first |D_k| cards of the new deck are D_k, the following |D_{k-1}| cards are D_{k-1} and so on). The internal order of each packet of cards D_i is unchanged by the operation. 



You have to obtain a sorted deck (i.e., a deck where the first card is 1, the second is 2 and so on) performing at most n operations. It can be proven that it is always possible to sort the deck performing at most n operations.

Examples of operation: The following are three examples of valid operations (on three decks with different sizes). 

  * If the deck is [3 6 2 1 4 5 7] (so 3 is the first card and 7 is the last card), we may apply the operation with k=4 and D_1=[3 6], D_2=[2 1 4], D_3=[5], D_4=[7]. Doing so, the deck becomes [7 5 2 1 4 3 6]. 
  * If the deck is [3 1 2], we may apply the operation with k=3 and D_1=[3], D_2=[1], D_3=[2]. Doing so, the deck becomes [2 1 3]. 
  * If the deck is [5 1 2 4 3 6], we may apply the operation with k=2 and D_1=[5 1], D_2=[2 4 3 6]. Doing so, the deck becomes [2 4 3 6 5 1]. 

Input

The first line of the input contains one integer n (1≤ n≤ 52) — the number of cards in the deck.

The second line contains n integers c_1, c_2, ..., c_n — the cards in the deck. The first card is c_1, the second is c_2 and so on.

It is guaranteed that for all i=1,...,n there is exactly one j∈\{1,...,n\} such that c_j = i.

Output

On the first line, print the number q of operations you perform (it must hold 0≤ q≤ n).

Then, print q lines, each describing one operation.

To describe an operation, print on a single line the number k of parts you are going to split the deck in, followed by the size of the k parts: |D_1|, |D_2|, ... , |D_k|. 

It must hold 2≤ k≤ n, and |D_i|≥ 1 for all i=1,...,k, and |D_1|+|D_2|+⋅⋅⋅ + |D_k| = n.

It can be proven that it is always possible to sort the deck performing at most n operations. If there are several ways to sort the deck you can output any of them.

Examples

Input


4
3 1 2 4


Output


2
3 1 2 1
2 1 3


Input


6
6 5 4 3 2 1


Output


1
6 1 1 1 1 1 1


Input


1
1


Output


0

Note

Explanation of the first testcase: Initially the deck is [3 1 2 4]. 

  * The first operation splits the deck as [(3) (1 2) (4)] and then transforms it into [4 1 2 3]. 
  * The second operation splits the deck as [(4) (1 2 3)] and then transforms it into [1 2 3 4]. 

Explanation of the second testcase: Initially the deck is [6 5 4 3 2 1]. 

  * The first (and only) operation splits the deck as [(6) (5) (4) (3) (2) (1)] and then transforms it into [1 2 3 4 5 6]. 

Solution (in JAVA):

Solution:
//created by Whiplash99
import java.io.*;
import java.util.*;
public class D
{
    private static StringBuilder sb;
    private static int solve(ArrayList<Integer> list, ArrayList<Integer> count)
    {
        int ops=0, size=list.size(), i, j=0;
        while (true)
        {
            boolean check=true;
            for(i=0;i<size-1;i++)
            {
                if(list.get(i)>list.get(i+1))
                    check=false;
            }
            if(check) return ops;
            ops++;

            for(i=0;i<size-1;i++)
            {
                if(list.get(i)==list.get(i+1)-count.get(i+1))
                {
                    count.set(i+1,count.get(i+1)+count.get(i));
                    list.remove(i);
                    count.remove(i);
                    size--; i--;
                }
            }

            outer: for(i=0;i<size;i++)
            {
                for(j=i+1;j<size;j++)
                {
                    if(list.get(i)-count.get(i)==list.get(j))
                        break outer;
                }
            }

            int c=0;
            List<Integer> lens=new ArrayList<>();
            ArrayList<Integer> tmp=new ArrayList<>();
            ArrayList<Integer> tmp2=new ArrayList<>();

            if(j<size-1)
            {
                c++; int z=0;
                for(int x=j+1;x<size;x++) z+=count.get(x);
                lens.add(z);

                for(int x=j+1;x<size;x++)
                {
                    tmp.add(list.get(x));
                    tmp2.add(count.get(x));
                }
            }

            c++;
            lens.add(count.get(j));
            tmp.add(list.get(j));
            tmp2.add(count.get(j));

            c++; int z=0;
            for(int x=i;x<j;x++) z+=count.get(x);
            lens.add(z);
            for(int x=i;x<j;x++)
            {
                tmp.add(list.get(x));
                tmp2.add(count.get(x));
            }

            if(i>0)
            {
                c++; z=0;

                for(int x=0;x<i;x++)
                {
                    tmp.add(list.get(x));
                    tmp2.add(count.get(x));
                    z+=count.get(x);
                }
                lens.add(z);
            }

            list=tmp;
            count=tmp2;

            Collections.reverse(lens);
            sb.append(c).append(" ");
            for(int zz:lens) sb.append(zz).append(" ");
            sb.append("\n");
        }
    }
    public static void main(String[] args) throws IOException
    {
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));

        int i,N;

        N=Integer.parseInt(br.readLine().trim());
        ArrayList<Integer> list=new ArrayList<>();
        String[] s=br.readLine().trim().split(" ");
        for(i=0;i<N;i++) list.add(Integer.parseInt(s[i]));

        ArrayList<Integer> count=new ArrayList<>();
        for(i=0;i<N;i++) count.add(1);

        sb=new StringBuilder();
        System.out.println(solve(list,count));
        System.out.println(sb);
    }
}