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 CodeForces;

import java.io.BufferedOutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.InputMismatchException;

public class GlobalRound11D {

	public static void solve() {
		int n = s.nextInt();
		int[] cards = new int[n];
		for(int i = 0; i < n; i++) {
			cards[i] = s.nextInt();
		}
		ArrayList<ArrayList<Integer>> ans = new ArrayList<ArrayList<Integer>>();
		while(true) {
			int min_idx = -1, my_big = -1;
			o : for(int i = 1; i <= n; i++) {
				for(int j = 0; j < n; j++) {
					if(cards[j] == i) {
						for(int k = 0; k < j; k++) {
							if(cards[k] == i + 1) {
								min_idx = j;
								my_big = k;
								break o;
							}
						}
						break;
					}
				}
			}
			if(min_idx == -1) {
				break;
			}
			// min_idx --> rangel, ranger
			int rangel = min_idx, ranger = min_idx;
			for(int i = min_idx - 1; i >= 0; i--) {
				if(cards[i] == cards[i + 1] - 1) {
					rangel--;
				}else {
					break;
				}
			}
			for(int i = min_idx + 1; i < n; i++) {
				if(cards[i] == cards[i - 1] + 1) {
					ranger++;
				}else {
					break;
				}
			}
			// my_big -> rl, rr
			int rl = my_big, rr = my_big;
			for(int i = my_big - 1; i >= 0; i--) {
				if(cards[i] == cards[i + 1] - 1) {
					rl--;
				}else {
					break;
				}
			}
			for(int i = my_big + 1; i < n; i++) {
				if(cards[i] == cards[i - 1] + 1) {
					rr++;
				}else {
					break;
				}
			}
			ArrayList<Integer> sizes = new ArrayList<Integer>();
			if(rl != 0) {
				sizes.add(rl);
			}
			sizes.add(rangel - rl);
			sizes.add(ranger - rangel + 1);
			if(ranger != n - 1) {
				sizes.add(n - 1 - ranger);
			}
			Reverse(cards, sizes);
			ans.add(sizes);
		}
		out.println(ans.size());
		for(ArrayList<Integer> x : ans) {
			out.print(x.size() + " ");
			for(Integer y : x) {
				out.print(y + " ");
			}
			out.println();
		}
	}
	
	public static void Reverse(int[] cards, ArrayList<Integer> sizes) {
		int start = 0;
		int end = cards.length - 1;
		int[] ans = new int[cards.length];
		for(Integer x : sizes) {
			int tempend = start + x - 1;
			while(tempend >= start) {
				ans[end] = cards[tempend];
				end--;
				tempend--;
			}
			start = start + x;
		}
		for(int i = 0; i < cards.length; i++) {
			cards[i] = ans[i];
		}
	}

	public static void main(String[] args) {
		new Thread(null, null, "Thread", 1 << 27) {
			public void run() {
				try {
					out = new PrintWriter(new BufferedOutputStream(System.out));
					s = new FastReader(System.in);
					solve();
					out.close();
				} catch (Exception e) {
					e.printStackTrace();
					System.exit(1);
				}
			}
		}.start();
	}

	public static PrintWriter out;
	public static FastReader s;

	public static class FastReader {

		private InputStream stream;
		private byte[] buf = new byte[4096];
		private int curChar, snumChars;

		public FastReader(InputStream stream) {
			this.stream = stream;
		}

		public int read() {
			if (snumChars == -1) {
				throw new InputMismatchException();
			}
			if (curChar >= snumChars) {
				curChar = 0;
				try {
					snumChars = stream.read(buf);
				} catch (IOException E) {
					throw new InputMismatchException();
				}
			}
			if (snumChars <= 0) {
				return -1;
			}
			return buf[curChar++];
		}

		public int nextInt() {
			int c = read();
			while (isSpaceChar(c)) {
				c = read();
			}
			int sgn = 1;
			if (c == '-') {
				sgn = -1;
				c = read();
			}
			int number = 0;
			do {
				number *= 10;
				number += c - '0';
				c = read();
			} while (!isSpaceChar(c));
			return number * sgn;
		}

		public long nextLong() {
			int c = read();
			while (isSpaceChar(c)) {
				c = read();
			}
			long sgn = 1;
			if (c == '-') {
				sgn = -1;
				c = read();
			}
			long number = 0;
			do {
				number *= 10L;
				number += (long) (c - '0');
				c = read();
			} while (!isSpaceChar(c));
			return number * sgn;
		}

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

		public long[] nextLongArray(int n) {
			long[] arr = new long[n];
			for (int i = 0; i < n; i++) {
				arr[i] = this.nextLong();
			}
			return arr;
		}

		public String next() {
			int c = read();
			while (isSpaceChar(c)) {
				c = read();
			}
			StringBuilder res = new StringBuilder();
			do {
				res.appendCodePoint(c);
				c = read();
			} while (!isSpaceChar(c));
			return res.toString();
		}

		public String nextLine() {
			int c = read();
			while (isSpaceChar(c)) {
				c = read();
			}
			StringBuilder res = new StringBuilder();
			do {
				res.appendCodePoint(c);
				c = read();
			} while (!isEndofLine(c));
			return res.toString();
		}

		public boolean isSpaceChar(int c) {
			return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
		}

		public boolean isEndofLine(int c) {
			return c == '\n' || c == '\r' || c == -1;
		}

	}

	
}