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:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.*;

public class Main
{
	public static void main(String[] args) {
		new Thread(null, new Runnable() {
			public void run() {
                solve();
            }
        }, "1", 1 << 26).start();
	}
	static void solve () {
		FastReader fr =new FastReader();	PrintWriter op =new PrintWriter(System.out);
 
 		int n =fr.nextInt() ,c[] =new int[n+1] ,d[] =new int[n+1] ,indx[] =new int[n+1] ,i ,j ,k ,l ,m ;

 		for (i =0 ; i<n ; ++i)	c[i+1] =fr.nextInt() ;

 		ArrayList<ArrayList<Integer>> list =new ArrayList<>() ;

 		ArrayList<Integer> decks =new ArrayList<>() ;

 		for (i =1 ; i<=n ; ++i) {
 			for (j =1 ; j<=n ; ++j)
 				indx[c[j]] =j ;

 			for (j =1 ; j<=n ; ++j) {
 				if (indx[j] != j)	break;
 			}

 			if (j>n)
 				break;
 			else {
 				decks.clear() ;

	 			if ((i&1) == 1) {
	 				if (indx[i] != 1)
	 					decks.add(indx[i]-1) ;

	 				decks.add(n-i-indx[i]+2) ;

	 				for (j =1 ; j<i ; ++j)
	 					decks.add(1) ;
	 			}
	 			else {
	 				for (j =1 ; j<i ; ++j)
	 					decks.add(1) ;

	 				decks.add(indx[i]-i+1) ;

	 				if (indx[i] != n)
	 					decks.add(n-indx[i]) ;
	 			}

	 			for (j =0 ,k =n+1 ,m =0 ; j<decks.size() ; ++j) {
	 				k -= decks.get(j) ;

	 				for (l =0 ; l<decks.get(j) ; ++l) {
	 					++m ;

	 					d[k+l] =c[m] ;
	 				}
	 			}

	 			for (j =1 ; j<=n ; ++j)
	 				c[j] =d[j] ;

	 			if (decks.size() > 1)
	 				list.add(new ArrayList<>(decks)) ;
	 		}
 		}

 		op.println(list.size()) ;

 		for (i =0 ; i<list.size() ; ++i) {
 			op.print(list.get(i).size()+" ") ;

 			for (j =0 ; j<list.get(i).size() ; ++j)
 				op.print(list.get(i).get(j)+" ") ;

 			op.println() ;
 		}

		op.flush();	op.close();
	}
	static class FastReader {
		BufferedReader br;
		StringTokenizer st;

		public FastReader() {
			br =new BufferedReader(new InputStreamReader(System.in));
		}

		String next() {
			while (st==null || (!st.hasMoreElements())) 
			{
				try
				{
					st =new StringTokenizer(br.readLine());
				}
				catch(IOException e)
				{
					e.printStackTrace();
				}
				
			}
			return st.nextToken();
		}

		String nextLine() {
			String str ="";

			try
			{
				str =br.readLine();
			}
			catch(IOException e)
			{
				e.printStackTrace();
			}

			return str;
		}

		int nextInt() {
			return Integer.parseInt(next());
		}

		long nextLong() {
			return Long.parseLong(next()) ;
		}
	}
}