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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.*;
import java.util.stream.Collectors;

public class SolutionD extends Thread {
    static class FastReader {
        BufferedReader br;
        StringTokenizer st;

        public FastReader() {
            br = new BufferedReader(new
                                            InputStreamReader(System.in));
        }

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

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

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

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

        String nextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
    }

    private static final FastReader scanner = new FastReader();
    private static final PrintWriter out = new PrintWriter(System.out);

    public static void main(String[] args) {
        new Thread(null, new SolutionD(), "Main", 1 << 26).start();
    }

    public void run() {
        int t = 1;
        for (int i = 0; i < t; i++) {
            solve();
        }
        out.close();
    }

    static class OutputLine {
        List<Integer> splitSize = new ArrayList<>();

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

    private static void solve() {
        int n = scanner.nextInt();
        int[] c = new int[n];
        for (int i = 0; i < n; i++) {
            c[i] = scanner.nextInt();
        }

        List<OutputLine> output = new ArrayList<>();

        while (true) {
            int wrongIndex = -1;
            for (int i = 0; i < n; i++) {
                if (i == 0 && c[i] != 1 || i > 0 && c[i] != c[i-1] + 1) {
                    wrongIndex = i;
                    break;
                }
            }
            if (wrongIndex == -1) {
                break;
            }

            int lowerNeighbor = -1;
            for (int i = 0; i < n; i++) {
                if (c[i] == c[wrongIndex] - 1) {
                    lowerNeighbor = i;
                    break;
                }
            }

            int lowerNeighborBeginn = lowerNeighbor;

            while (c[lowerNeighborBeginn] == c[lowerNeighborBeginn-1] + 1) {
                lowerNeighborBeginn--;
            }

            OutputLine line = new OutputLine();

            int[] newC = new int[n];

            if (wrongIndex > 0) {
                line.splitSize.add(wrongIndex);
            }
            line.splitSize.add(lowerNeighborBeginn - wrongIndex);
            line.splitSize.add(lowerNeighbor - lowerNeighborBeginn + 1);

            if (lowerNeighbor != n - 1) {
                line.splitSize.add(n - lowerNeighbor - 1);
            }

            int pointerOld = n;
            int pointerNew = 0;
            for (int i = line.splitSize.size() - 1; i >= 0; i--) {
                pointerOld -= line.splitSize.get(i);
                for (int j = 0; j < line.splitSize.get(i); j++) {
                    newC[pointerNew+j] = c[pointerOld+j];
                }
                pointerNew += line.splitSize.get(i);
            }
            c = newC;
            output.add(line);
        }

        System.out.println(output.size());
        output.forEach(System.out::println);
    }
}