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 four {

	public static void main(String[] args) throws IOException, FileNotFoundException {
		BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
		//BufferedReader in = new BufferedReader(new FileReader("four"));

		int n = Integer.parseInt(in.readLine());
		int[] arr = new int[n+1];
		StringTokenizer st = new StringTokenizer(in.readLine());
		for (int i=1; i<=n; i++) {
			arr[i] = Integer.parseInt(st.nextToken());
		}
		
		ArrayList<ArrayList<Integer>> moves = new ArrayList<>();
		boolean front = false;
		for (int i=n; i>=1; i--) {
			if (n%2 != 0 && i==1) break;
			ArrayList<Integer> cur = new ArrayList<Integer>();
			int[] temp = new int[n+1];
			if (front) {
				for (int j=n; j>i; j--) {
					cur.add(1);
					temp[j] = j;
				}
				
				int j= (n-i+1);
				for (; j<n; j++) {
					if (arr[j] == i) break;
				}
				
				if (j - (n-i) > 0) cur.add(j - (n-i));
				if (n - j > 0) cur.add(n - j);
	
				int p=1;
				for (int k=j+1; k<= n; k++) {
					temp[p] = arr[k];
					++p;
				}

				for (int k=n-i+1; k<=j; k++) {
					temp[p] = arr[k];
					p++;
				}
			}
			else {
				
				int j=1;
				for (; j<=n; j++ ) {
					if (arr[j] == i) break;
				}
				
				if (j-1 > 0) cur.add(j-1);
				for (int k=j-1; k>=0; k--) {
					temp[n+(k-j + 1)] = arr[k];
				}
				
				int prev = j;
				for (; j<=n; j++) {
					if (arr[j] > i) break;
				}
				j--;
				
				if (j - prev + 1 > 0) cur.add(j - prev+1);
				
				int p=1;
				for (int k=n; k>j; k--) {
					cur.add(1);
					temp[p] = arr[k];
					p++;
				}
				
				for (int k=prev; k<=j; k++) {
					temp[p] = arr[k];
					p++;
				}
			}
			
			arr = temp;
			if (cur.size() != 1) moves.add(cur);
			front = !front;
		}
		
		System.out.println(moves.size());
		for (int i=0; i<moves.size(); i++) {
			System.out.print(moves.get(i).size() + " ");
			for (int j=0; j<moves.get(i).size(); j++) {
				System.out.print(moves.get(i).get(j) + " ");
			}
			System.out.println();
		}

	}
}