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.*;
import java.util.*;

public class D {

	public static void main(String[] args) {
		FastScanner sc = new FastScanner();
		int n = sc.nextInt();
		int[] arr = new int[n];
		StringBuilder sbres = new StringBuilder();
		for(int i = 0; i < n; i++) {
			arr[i] = sc.nextInt()-1;
		}
		int high = n/2, low = n/2-1;
		int count = 0;
		for(int iter = 0; iter < n; iter++) {
			ArrayList<Integer> vals = new ArrayList<>();
			if(iter % 2 == 0) { // right to left
				int x = 0;
				for(int i = 0; i < n; i++) {
					if(arr[i] == high) {
						x = i; break;
					}
				}
				vals.add(x);
				vals.add(n - iter - x);
				vals.add(iter);
				int[] newarr = new int[n];
				int k = 0;
				for(int i = n-iter; i < n; i++) {
					newarr[k++] = arr[i];
				}
				for(int i = x; i < n-iter; i++) {
					newarr[k++] = arr[i];
				}
				for(int i = 0; i < x; i++) {
					newarr[k++] = arr[i];
				}
				arr = newarr;
				high++;
			}
			else { // left to right
				int x = 0;
				for(int i = 0; i < n; i++) {
					if(arr[i] == low) {
						x = i; break;
					}
				}
				vals.add(iter);
				vals.add(x - iter + 1);
				vals.add(n - x - 1);
				int[] newarr = new int[n];
				int k = 0;
				for(int i = x+1; i < n; i++) {
					newarr[k++] = arr[i];
				}
				for(int i = iter; i <= x; i++) {
					newarr[k++] = arr[i];
				}
				for(int i = 0; i < iter; i++) {
					newarr[k++] = arr[i];
				}
				arr = newarr;
				low--;
			}
			String s = convert(vals);
			if(s != null) {
				sbres.append(s+"\n"); count++;
			}
		}
		PrintWriter pw = new PrintWriter(System.out);
		pw.print(count+"\n");
		pw.print(sbres.toString());
		pw.flush();
	}
	static String convert(ArrayList<Integer> list) {
		StringBuilder sb = new StringBuilder();
		int c = 0;
		for(int i = 0; i < list.size(); i++) {
			if(list.get(i) > 0) {
				sb.append(" "+list.get(i));
				c++;
			}
		}
		if(c == 1) return null;
		return c+sb.toString();
	}
	
	static class FastScanner {
        public BufferedReader reader;
        public StringTokenizer tokenizer;
        public FastScanner() {
            reader = new BufferedReader(new InputStreamReader(System.in), 32768);
            tokenizer = null;
        }
        public String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }
        public int nextInt() {
            return Integer.parseInt(next());
        }
        public long nextLong() {
        	return Long.parseLong(next());
        }
        public double nextDouble() {
        	return Double.parseDouble(next());
        }
        public String nextLine() {
        	try {
        		return reader.readLine();
        	} catch(IOException e) {
        		throw new RuntimeException(e);
        	}
        }
    }

}