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 C1427 {

	static int[] arr;

	public static void performOp(ArrayList<Integer> size) {
		int curIdx = 0;
		ArrayList<int[]> parts = new ArrayList<int[]>();
		for (int x : size) {
			int[] part = new int[x];
			for (int i = 0; i < x; i++) {
				part[i] = arr[curIdx++];
			}
			parts.add(part);
		}
		curIdx = 0;
		for (int i = parts.size() - 1; i >= 0; i--) {
			int[] part = parts.get(i);
			for (int j = 0; j < part.length; j++) {
				arr[curIdx++] = part[j];
			}
		}
	}

	public static void findOp(int idxSmall, int idxBig) {
		ArrayList<Integer> op = new ArrayList<Integer>();
		if (idxBig != 0) {
			op.add(idxBig);
		}
		int mid = idxBig;
		while (arr[mid + 1] == arr[mid] + 1) {
			mid++;
		}
		op.add(mid - idxBig + 1);
		op.add(idxSmall - mid);
		if (idxSmall != arr.length - 1) {
			op.add(arr.length - 1 - idxSmall);
		}
		ans.add(op);
		performOp(op);
	}

	static ArrayList<ArrayList<Integer>> ans = new ArrayList<ArrayList<Integer>>();

	public static void main(String[] args) throws IOException {
		Scanner sc = new Scanner(System.in);
		PrintWriter pw = new PrintWriter(System.out);
		int n = sc.nextInt();
		arr = sc.nextIntArr(n);
		for (int k = 0; k < n; k++) {
			for (int i = 1; i < n; i++) {
				int idxI = -1, idxNxt = -1;
				for (int j = 0; j < arr.length; j++) {
					if (arr[j] == i) {
						idxI = j;
					} else if (arr[j] == i + 1) {
						idxNxt = j;
					}
				}
				if (idxNxt < idxI) {
					findOp(idxI, idxNxt);
					break;
				}
			}
		}
//		System.out.println(Arrays.toString(arr));
		pw.println(ans.size());
		for (ArrayList<Integer> op : ans) {
			pw.print(op.size());
			for (int x : op) {
				pw.print(" " + x);
			}
			pw.println();
		}
		pw.close();
	}

	static class Scanner {
		BufferedReader br;
		StringTokenizer st;

		public Scanner(InputStream s) {
			br = new BufferedReader(new InputStreamReader(s));
		}

		public Scanner(FileReader f) {
			br = new BufferedReader(f);
		}

		public String next() throws IOException {
			while (st == null || !st.hasMoreTokens())
				st = new StringTokenizer(br.readLine());
			return st.nextToken();
		}

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

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

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

		public int[] nextIntArr(int n) throws IOException {
			int[] arr = new int[n];
			for (int i = 0; i < n; i++) {
				arr[i] = Integer.parseInt(next());
			}
			return arr;
		}

	}

}
