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.ByteArrayInputStream;
import java.io.File;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;

public class D {
  public static void main(String[] args) throws Exception {
    new D().go();
  }

  int USE_FILE = 0;
  String INPUT = "6\r\n" +
      "6 5 4 3 2 1\r\n" +
      "";

  void solve(int caseNr) {
    int n = in.nextInt();
    int[] a = next_int_array(n);

    int cost = 0;
    List<List<Integer>> moves = new ArrayList<>();

    while(!isSorted(a)) {
      cost++;

      ArrayList<Integer> cuts = new ArrayList<>();
      int ind = n;
      while(ind == a[ind-1]) {
        ind--;
      }
      if(ind<n) cuts.add(ind);
      ind--;
      int right=a[ind];
      while(a[ind-1] == a[ind]-1) {
        ind--;
      }
      cuts.add(0, ind);
      ind--;
      while(a[ind] != right+1) {
        ind--;
      }
      if(ind>0) cuts.add(0, ind);

      List<Integer> move = new ArrayList<>();
      move.add(cuts.size()+1);
      move.add(cuts.get(0));
      for (int i = 1; i < cuts.size(); i++) {
        move.add(cuts.get(i)-cuts.get(i-1));
      }
      move.add(n-cuts.get(cuts.size()-1));
      moves.add(move);
      a = shuffle(a, cuts);
    }

    System.out.println(cost);
    for (List<Integer> list : moves) {
      printList(list);
    }
  }

  int[] shuffle(int[] a, List<Integer> cuts) {
    int n = a.length;
    int[] b = new int[n];

    int i1=0;
    for (Integer i2 : cuts) {
      int d=0;
      for (int i = i1; i < i2; i++) {
        b[n-i2+d] = a[i];
        d++;
      }
      i1=i2;
    }
    int d=0;
    for (int i = i1; i < n; i++) {
      b[d++] = a[i];
    }
    return b;
  }

  boolean isSorted(int[] a) {
    for (int i = 0; i < a.length; i++) {
      if (a[i] != i+1) return false;
    }
    return true;
  }



  // ---------------------------------------------------------------------------------------------- //

  void testCases() {
    int t = 1;
    for (int i=1; i<=t; i++) {
      solve(i);
    }
    output_writer.flush();
  }

  // ---------------------------------------------------------------------------------------------- //
  // ---------------------------------------------------------------------------------------------- //

  static boolean LOCAL = java.lang.System.getProperty("ONLINE_JUDGE") == null;
  Scanner in;
  PrintWriter output_writer = new PrintWriter(java.lang.System.out);

  // the output musn't flush automatic or we time out
  static class System {
    static PrintWriter out = null;
  }

  void go() throws Exception {
    System.out = output_writer;

    if (LOCAL && USE_FILE == 0) {
      //  if not set expects input from terminal
      java.lang.System.setIn(new ByteArrayInputStream(INPUT.getBytes()));
      in = new Scanner(java.lang.System.in);
    }
    if (LOCAL && USE_FILE > 0) {
      String filename = String.format("in/cf%d.txt", USE_FILE);
      File inputFile = new File(filename);
      in = new Scanner(inputFile);
      output_writer.printf("[%s]\n", filename);
      output_writer.flush();
    }
    if (!LOCAL) {
      in = new Scanner(java.lang.System.in);
    }


    long startTime = java.lang.System.currentTimeMillis();
    testCases();
    if (LOCAL) {
      java.lang.System.out.printf("[%dms]\n", java.lang.System.currentTimeMillis()-startTime);
    }
    in.close();
  }

  // ---------------------------------------------------------------------------------------------- //

  private int[] next_int_array(int n) {
    int[] a = new int[n];
    for (int i=0; i<n; i++) {
      a[i] = in.nextInt();
    }
    return a;
  }
  private void printList(List<Integer> a) {
    for (int i = 0; i <= a.size()-1; i++) {
      System.out.print(a.get(i) + " ");
    }
    System.out.println();
  }
}





