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:
//package com.cf.global11;

import java.io.BufferedReader;
import java.io.File;
import java.io.FileNotFoundException;
import java.io.FileReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.List;
import java.util.StringTokenizer;
import java.util.stream.Collectors;
import java.util.stream.IntStream;

public class D {

	FastScanner in;
	PrintWriter out;

	public static void main(String[] args) {
		new D().solve();
	}

	private void solve() {
		in = new FastScanner(System.in);
		out = new PrintWriter(System.out);

		solveCase();

		out.close();
	}

	private void solveCase() {
		int n = in.nextInt();

		List<Integer> cards = IntStream.range(0, n).map(i -> in.nextInt()).boxed().collect(Collectors.toList());

		List<Operation> operations = new ArrayList<>();

		for (int i = 0;2 * i < n - 1;i++) if (cards.get(i) != i + 1 || cards.get(n - 1 - i) != n - i) {
			List<Operation> nextOperations =
					sortingOperationsFor(cards, i)
						.stream()
						.filter(x -> x.sizes.size() > 1)
						.collect(Collectors.toList());
			operations.addAll(nextOperations);
			for (Operation nextOp : nextOperations) {
				cards = nextOp.operateOn(cards);
			}
		}

		out.println(operations.size());
		for (Operation operation : operations) {
			out.println(operation);
		}
	}

	private List<Operation> sortingOperationsFor(List<Integer> cards, int i) {
		int a1 = cards.indexOf(i + 1);
		int a2 = cards.indexOf(cards.size() - i);
		if (a1 < a2) {
			return List.of(
					new Operation(List.of(i, a1 - i, 1, a2 - a1 - 1, 1, cards.size() - 1 - a2 - i, i)),
					new Operation(List.of(i, cards.size() - a2 - i, a2 - a1 - 1, a1 + 1 - i, i)));
		} else {
			return List.of(
					new Operation(List.of(i, a2 + 1 - i, a1 - a2 - 1, cards.size() - a1 - i, i)),
					new Operation(List.of(i, cards.size() - 2 * i, i)));
		}
	}

	static class FastScanner {

		BufferedReader br;
		StringTokenizer st;

		public FastScanner(File f) {
			try {
				br = new BufferedReader(new FileReader(f));
			} catch (FileNotFoundException e) {
				e.printStackTrace();
			}
		}

		public FastScanner(InputStream f) {
			br = new BufferedReader(new InputStreamReader(f));
		}

		String next() {
			while (st == null || !st.hasMoreTokens()) {
				String s = null;
				try {
					s = br.readLine();
				} catch (IOException e) {
					e.printStackTrace();
				}
				if (s == null)
					return null;
				st = new StringTokenizer(s);
			}
			return st.nextToken();
		}

		boolean hasMoreTokens() {
			while (st == null || !st.hasMoreTokens()) {
				String s = null;
				try {
					s = br.readLine();
				} catch (IOException e) {
					e.printStackTrace();
				}
				if (s == null)
					return false;
				st = new StringTokenizer(s);
			}
			return true;
		}

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

		long nextLong() {
			return Long.parseLong(next());
		}

		double nextDouble() {
			return Double.parseDouble(next());
		}
	}

	private static class Celebrity {
		int t, x, y;

		public Celebrity(int t, int x, int y) {
			this.t = t;
			this.x = x;
			this.y = y;
		}
	}

	private static class Operation {

		public final List<Integer> sizes;

		private Operation(List<Integer> sizes) {
			this.sizes = sizes.stream().filter(s -> s > 0).collect(Collectors.toList());
		}

		public List<Integer> operateOn(List<Integer> cards) {
			if (sizes.stream().mapToInt(x -> x).sum() != cards.size()) throw new RuntimeException("bug!");

			List<Integer> newCards = new ArrayList<>(cards);
			int alreadyPut = 0;
			for (int size : sizes) {
				for (int i = 0; i < size; i++) {
					newCards.set(cards.size() - alreadyPut - size + i, cards.get(alreadyPut + i)); //TODO: check it out
				}
				alreadyPut += size;
			}

			return newCards;
		}

		@Override
		public String toString() {
			return sizes.size() + " " +
					sizes.stream()
					.map(String::valueOf).collect(Collectors.joining(" "));
		}
	}
}