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.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.List;
import java.util.Arrays;
import java.util.StringTokenizer;
import java.io.IOException;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 *
 * @author lucasr
 */
public class Main {
	public static void main(String[] args) {
		InputStream inputStream = System.in;
		OutputStream outputStream = System.out;
		MyScanner in = new MyScanner(inputStream);
		PrintWriter out = new PrintWriter(outputStream);
		DUnshufflingADeck solver = new DUnshufflingADeck();
		solver.solve(1, in, out);
		out.close();
	}

	static class DUnshufflingADeck {
		public void solve(int testNumber, MyScanner sc, PrintWriter out) {
			int n = sc.nextInt();
			int[] elems = sc.nextIntArray(n);
			List<int[]> ret = new ArrayList<>();
			int[] tmp;
			while ((tmp = process(elems)) != null) {
				ret.add(tmp);
				elems = transform(elems, tmp);
			}
			out.println(ret.size());
			for (int[] ints : ret) {
				out.print(ints.length + " ");
				for (int val : ints) {
					out.print(val + " ");
				}
				out.println();
			}
		}

		static int[] process(int[] elems) {
			int n = elems.length;
			for (int i = 0; i < n; i++) {
				for (int j = i + 1; j < n; j++)
					if (elems[i] == elems[j] + 1) {
						for (int k = i; k < j; k++) {
							if (elems[k] + 1 != elems[k + 1]) {
								int[] ret = new int[4];
								int size = 0;
								if (i > 0) {
									ret[size++] = i;
								}
								ret[size++] = k - i + 1;
								ret[size++] = j - k;
								if (n > j + 1) {
									ret[size++] = n - j - 1;
								}
								return Arrays.copyOf(ret, size);
							}
						}

					}
			}
			return null;
		}

		static int[] transform(int[] elems, int[] parts) {
			int n = elems.length;
			int[] fArr = new int[parts.length];
			for (int i = 1; i < fArr.length; i++) {
				fArr[i] = fArr[i - 1] + parts[i - 1];
			}
			int[] ret = new int[n];
			int from = 0;
			for (int i = parts.length - 1; i >= 0; i--) {
				for (int j = 0; j < parts[i]; j++) {
					ret[from + j] = elems[fArr[i] + j];
				}
				from += parts[i];
			}
			return ret;
		}

	}

	static class MyScanner {
		private BufferedReader br;
		private StringTokenizer tokenizer;

		public MyScanner(InputStream is) {
			br = new BufferedReader(new InputStreamReader(is));
		}

		public String next() {
			while (tokenizer == null || !tokenizer.hasMoreTokens()) {
				try {
					tokenizer = new StringTokenizer(br.readLine());
				} catch (IOException e) {
					throw new RuntimeException(e);
				}
			}
			return tokenizer.nextToken();
		}

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

		public int[] nextIntArray(int n) {
			int[] ret = new int[n];
			for (int i = 0; i < n; i++) {
				ret[i] = nextInt();
			}
			return ret;
		}

	}
}

