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]. 
import java.util.*;
import java.io.*;

public class D1427{

	public static boolean sorted(int[] x){
		boolean sorted = true;
		for(int i = 0; i < x.length-1; i++){
			if(x[i] != x[i + 1]-1){
				sorted = false;
				break;
			}
		}
		return sorted;
	}

	public static ArrayList<Integer> extract(int[] x, int a, int b){
		ArrayList<Integer> alist = new ArrayList<Integer>();
		for(int i = a; i < b; i++){
			alist.add(x[i]);
		}
		return alist;
	}

	public static void insert(int[] x, ArrayList<Integer> a, int count){
		for(int i = 0; i < a.size(); i++){
			x[i + count] = a.get(i);
		}
	}

	public static void main(String[] args) throws Exception{
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		
		int n = Integer.parseInt(br.readLine());

		int[] x = new int[n];
		StringTokenizer st = new StringTokenizer(br.readLine());
		for(int i = 0; i < n; i++){
			x[i] = Integer.parseInt(st.nextToken()) - 1;
		}

		if(sorted(x)){
			System.out.println("0");
			return;
		}

		ArrayList<List<Integer>> moves = new ArrayList<List<Integer>>();

		outer:
		for(int i = 0; i < n; i++){
			if(sorted(x))
				break;
			for(int j = 0; j < n - 1; j++){
				int indexj = -1;
				int indexj2 = -1;
				for(int k = 0; k < n; k++){
					if(x[k] == j){
						indexj = k;
					}
					if(x[k] == j + 1){
						indexj2 = k;
					}
				}
				if(indexj2 > indexj)
					continue;
				// a b c d
				int a = 0;
				int b = indexj2;
				int c = indexj2 + 1;
				int d = indexj + 1;
				while(x[c] == x[c - 1] + 1){
					c++;
				}
				TreeSet<Integer> indexes = new TreeSet<Integer>();
				indexes.add(a);
				indexes.add(b);
				indexes.add(c);
				if(d != n )
					indexes.add(d);
				moves.add(new ArrayList<Integer>(indexes));
				ArrayList<Integer> alist = extract(x, a, b);
				ArrayList<Integer> blist = extract(x, b, c);
				ArrayList<Integer> clist = extract(x, c, d);
				ArrayList<Integer> dlist = extract(x, d, n);
				int count = 0;
				insert(x, dlist, count);
				count += dlist.size();
				insert(x, clist, count);
				count += clist.size();
				insert(x, blist, count);
				count += blist.size();
				insert(x, alist, count);
				count += alist.size();

				break;
			}
		}

		// System.out.println(Arrays.toString(x));

		StringBuilder sb = new StringBuilder();
		sb.append(moves.size() + "\n");
		for(int i = 0; i < moves.size(); i++){
			sb.append(moves.get(i).size() + " ");
			moves.get(i).add(n);
			for(int j = 0; j < moves.get(i).size() - 1; j++){
				int diff = moves.get(i).get(j + 1) - moves.get(i).get(j);
				sb.append(diff + " ");
			}
			sb.append("\n");
		}

		System.out.print(sb);

	}

}