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:
import java.io.*;
import java.util.*;

public class Codeforces
{
    public static void main(String args[])throws Exception
    {
        BufferedReader bu=new BufferedReader(new InputStreamReader(System.in));
        StringBuilder sb=new StringBuilder();
        int n=Integer.parseInt(bu.readLine());
        String s[]=bu.readLine().split(" ");
        int i,a[]=new int[n+1],tem[]=new int[n+1];
        for(i=1;i<=n;i++)
        a[i]=Integer.parseInt(s[i-1]);

        ArrayList<ArrayList<Integer>> ans=new ArrayList<>();
        while(unsorted(a)==true)
        {
            int ti=0,tj=0,t,j,l;
            for(i=1;i<=n;i++)
            for(j=i+1;j<=n;++j)
            if(a[i]==a[j]+1)
            {
                ti=i; tj=j;
                break;
            }
            t=tj;
            ArrayList<Integer> cur=new ArrayList<>();
            while(ti<t-1 && a[t-1]==a[t]-1) --t;
            for(i=1,j=n,l=1;i<=n;++i)
            if(i==ti-1 || i==t-1 || i==tj || i==n)
            {
                    cur.add(i-l+1);
                    for(int k=i;k>=l;--k,--j) tem[j]=a[k];
                    l=i+1;
            }
            for(i=1;i<=n;++i) a[i]=tem[i];
            ans.add(cur);
        }

        sb.append(ans.size()+"\n");
        for(i=0; i<ans.size();i++)
        {
            sb.append(ans.get(i).size()+" ");
            for(int j=0;j<ans.get(i).size();j++)
                sb.append(ans.get(i).get(j)+" ");
            sb.append("\n");
        }
        System.out.print(sb);
    }

    static boolean unsorted(int a[])
    {
        int i,n=a.length;
        for(i=1;i<n-1;i++)
        if(a[i]>a[i+1]) return true;
        return false;
    }
}