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.util.*;
import java.io.*;
public class EdD {
	static long num = 998244353;
	public static void main(String[] args) throws Exception{
		

		// TODO Auto-generated method stub
 		BufferedReader bf = new BufferedReader(new InputStreamReader(System.in));
 		PrintWriter out = new PrintWriter(System.out);
//	 		String input1 = bf.readLine().trim();
//	 		String input2 = bf.readLine().trim();
 			int n = Integer.parseInt(bf.readLine());
	 		ArrayList<Integer> array = new ArrayList<Integer>();
	 		StringTokenizer st2 = new StringTokenizer(bf.readLine());
	 		for(int j=0;j<n;j++){
	 			array.add(Integer.parseInt(st2.nextToken()));
	 		}
	 		ArrayList<Answers> ans = new ArrayList<Answers>();
	 		int[] groups = new int[n+1];
	 		for(int j =1;j<=n;j++){
	 			groups[j] = 1;
	 		}
	 		
	 		while(groups[1] != n && !isSorted(array)){
	 			int k = next(n, array.get(array.size()-1));
	 			int index = 0;
	 			for(int j = 0;j<array.size();j++){
	 				if (array.get(j) == k){
	 					index = j;
	 					break;
	 					
	 				}
	 			}
	 			int[] newGroups = new int[n+1];
	 			for(int j =0 ;j<=n;j++){
	 				newGroups[j] = groups[j];
	 			}
	 			for(int ts = n-1;ts>=n-groups[array.get(array.size()-1)];ts--){
	 				newGroups[array.get(ts)]+=groups[array.get(index)];
	 			}
	 			for(int ts = index;ts<index+groups[array.get(index)];ts++){
	 				newGroups[array.get(ts)] = newGroups[array.get(array.size()-1)];
	 			}

	 			if (index == 0){
	 				ArrayList<Integer> temp = new ArrayList<Integer>();
	 				temp.add(n-groups[array.get(array.size()-1)]);
	 				temp.add(groups[array.get(array.size()-1)]);
	 				ans.add(new Answers(2, temp));
	 				ArrayList<Integer> newArray = new ArrayList<Integer>();
	 				for(int j = array.size()-groups[array.get(array.size()-1)];j<array.size();j++)
	 					newArray.add(array.get(j));
	 				for(int j =0 ;j<array.size()-groups[array.get(array.size()-1)];j++)
	 					newArray.add(array.get(j));
	 				array.clear();
	 				array.addAll(newArray);
	 			}
	 			else{
	 				ArrayList<Integer> temp = new ArrayList<Integer>();
	 				temp.add(index);
	 				temp.add(n-index-groups[array.get(array.size()-1)]);
	 				temp.add(groups[array.get(array.size()-1)]);
	 				ans.add(new Answers(3, temp));
	 				ArrayList<Integer> newArray = new ArrayList<Integer>();
	 				for(int j = array.size()-groups[array.get(array.size()-1)];j<array.size();j++)
	 					newArray.add(array.get(j));
	 				for(int j =index ;j<array.size()-groups[array.get(array.size()-1)];j++)
	 					newArray.add(array.get(j));
	 				for(int j = 0;j<index;j++)
	 					newArray.add(array.get(j));
	 				array.clear();
	 				array.addAll(newArray);
	 			}
	 			for(int j =0 ;j<=n;j++){
	 				groups[j] = newGroups[j];
	 			}
	 			
	 		}
	 		
	 		
	 		if (isSorted(array)){
		 		out.println(ans.size());
		 		for(Answers j : ans){
		 			out.print(j.size + " ");
		 			for(int k : j.a){
		 				out.print(k + " ");
		 			}
		 			out.println();
		 		}
	 		}
	 		else{
	 			out.println(ans.size()+1);
	 			for(Answers j : ans){
		 			out.print(j.size + " ");
		 			for(int k : j.a){
		 				out.print(k + " ");
		 			}
		 			out.println();
		 		}
	 			int index =0 ;
	 			for(int j = 0;j<n;j++){
	 				if (array.get(j) == 1){
	 					index = j;
	 					break;
	 				}
	 			}
	 			out.println("2 " + index + " " + (n-index)); 
	 		}
	 		
	 		
	 		
	 		
 		
 		
 		
 		out.println();
	 		
 		out.close();
 		
 		
 		
 	}
	static class Answers{
		private int size;
		private ArrayList<Integer> a;
		public Answers(int size, ArrayList<Integer> a){
			this.a = a;
			this.size = size;
		}
	}
	public static int next(int n, int k){
		if (k+1 <=n)
			return k+1;
		else
			return k+1-n;
	}
	public static boolean isSorted(ArrayList<Integer> a){
		for(int j =0;j<a.size()-1;j++){
			if (a.get(j) > a.get(j+1))
				return false;
		}
		return true;
	}
}
 	
 
//StringJoiner sj = new StringJoiner(" "); 
//sj.add(strings)
//sj.toString() gives string of those stuff w spaces or whatever that sequence is