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]. 
/*
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Don't let that studying and grades consume you
Just live life young
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That means I'm on top of my game and it don't stop
til my hip don't hop anymore
https://www.a2oj.com/Ladder16.html
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Shining through the city with a little funk and soul
*/
import java.util.*;
import java.io.*;
import java.math.*;

   public class D
   {
      public static void main(String hi[]) throws Exception
      {
         BufferedReader infile = new BufferedReader(new InputStreamReader(System.in));
         StringTokenizer st = new StringTokenizer(infile.readLine());
         int N = Integer.parseInt(st.nextToken());
         int[] arr = new int[N];
         st = new StringTokenizer(infile.readLine());
         for(int i=0; i < N; i++)
            arr[i] = Integer.parseInt(st.nextToken());
         ArrayList<Integer>[] operations = new ArrayList[N];
         int turns;
         matcha:for(turns=0; turns < N; turns++)
         {
            if(sorted(arr))
               break;
            boolean[] seen = new boolean[N+1];
            for(int a=N-1; a >= 0; a--)
            {
               seen[arr[a]] = true;
               if(arr[a] < N && !seen[arr[a]+1])
               {
                  int b = -1;
                  for(int i=0; i < N; i++)
                     if(arr[i] == arr[a]+1)
                     {
                        b = i;
                        break;
                     }
                  int actuala = a;
                  while(actuala > 0 && arr[actuala-1] == arr[actuala]-1)
                     actuala--;
                  ArrayList<Integer> ls = operate(arr, actuala, a, b);
                  operations[turns] = ls;
                  continue matcha;
               }
            }
         }
         StringBuilder sb = new StringBuilder();
         sb.append(turns+"\n");
         for(int i=0; i < turns; i++)
         {
            sb.append(operations[i].size()+" ");
            for(int x: operations[i])
               sb.append(x+" ");
            sb.append("\n");
         }
         System.out.print(sb);
      }
      public static ArrayList<Integer> operate(int[] arr, int lefta, int righta, int b)
      {
         int N = arr.length;
         ArrayList<Integer> ls = new ArrayList<Integer>();
         int mid1 = lefta-b;
         int mid2 = righta-lefta+1;
         int end = N-righta-1;
         if(b > 0)
            ls.add(b);
         ls.add(mid1);
         ls.add(mid2);
         if(end > 0)
            ls.add(end);
         int[] next = new int[N];
         Collections.reverse(ls);
         int right = N-1;
         int lol = 0;
         for(int len: ls)
         {
            int dex = right-len+1;
            while(dex <= right)
               next[lol++] = arr[dex++];
            right -= len;
         }
         for(int i=0; i < N; i++)
            arr[i] = next[i];
         Collections.reverse(ls);
         return ls;
      }
      public static boolean sorted(int[] arr)
      {
         for(int i=0; i < arr.length; i++)
            if(arr[i] != i+1)
               return false;
         return true;
      }
   }