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

public class Main {
    public static void main(String args[]) {new Main().run();}

    FastReader in = new FastReader();
    PrintWriter out = new PrintWriter(System.out);
    void run(){
        work();
        out.flush();
    }
    long mod=1000000007;
    long gcd(long a,long b) {
        return a==0?b:gcd(b%a,a);
    }

    void work() {
        int n=ni();
        int[] A=nia(n);
        boolean isRev=true;
        List<List<Integer>> ret=new ArrayList<>();
        for(int cur=n;cur>=2;cur--){
            int idx=getIdx(A,cur);
            int[] B=new int[n];
            List<Integer> list=new ArrayList<>();
            if(isRev){
                //0->n-cur-1  n-cur--idx  idx+1--n-1
                int i=0,p=n-1;
                for(int j=0;j<n-cur;j++){
                    list.add(1);
                }
                if(idx-(n-cur)+1>0)list.add(idx-(n-cur)+1);
                if(n-1-idx>0)list.add(n-1-idx);
                for(int j=0;j<n-cur;j++){
                    B[p--]=A[j];
                }
                for(int j=idx;j>=n-cur;j--){
                    B[p--]=A[j];
                }
                for(int j=n-1;j>=idx+1;j--){
                    B[p--]=A[j];
                }
            }else{
               //0 --- idx-1  idx---cur-1  cur->n-1
                int i=0,p=n-1;
                if(idx>0)list.add(idx);
                if(cur-idx>0)list.add(cur-idx);
                for(int j=cur;j<n;j++){
                    list.add(1);
                }

                for(int j=idx-1;j>=0;j--){
                    B[p--]=A[j];
                }
                for(int j=cur-1;j>=idx;j--){
                    B[p--]=A[j];
                }
                for(int j=cur;j<n;j++){
                    B[p--]=A[j];
                }
            }
            if(list.size()>1)ret.add(list);
            A=B;
            isRev=!isRev;
        }
        if(isRev){
            List<Integer> list=new ArrayList<>();
            for(int i=0;i<n;i++){
                list.add(1);
            }
            if(list.size()>1)ret.add(list);
        }
        out.println(ret.size());
        for(List<Integer> l:ret){
            out.print(l.size()+" ");
            for(int i=0;i<l.size();i++){
                out.print(l.get(i));
                out.print(" ");
            }
            out.println();
        }
    }

    private int getIdx(int[] A, int cur) {
        for(int i=0;i<A.length;i++){
            if(A[i]==cur)return i;
        }
        return -1;
    }

    private ArrayList<long[]>[] ngw(int n, int m) {
        ArrayList<long[]>[] graph=(ArrayList<long[]>[])new ArrayList[n];
        for(int i=0;i<n;i++) {
            graph[i]=new ArrayList<>();
        }
        for(int i=1;i<=m;i++) {
            long s=in.nextLong()-1,e=in.nextLong()-1,w=in.nextLong();
            graph[(int)s].add(new long[] {e,w,i});
            graph[(int)e].add(new long[] {s,w});
        }
        return graph;
    }

    private int ni() {
        return in.nextInt();
    }

    private long nl() {
        return in.nextLong();
    }

    private String ns() {
        return in.next();
    }

    private long[] na(int n) {
        long[] A=new long[n];
        for(int i=0;i<n;i++) {
            A[i]=in.nextLong();
        }
        return A;
    }
    private int[] nia(int n) {
        int[] A=new int[n];
        for(int i=0;i<n;i++) {
            A[i]=in.nextInt();
        }
        return A;
    }
}

class FastReader
{
    BufferedReader br;
    StringTokenizer st;

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


    public String next()
    {
        while(st==null || !st.hasMoreElements())//回车，空行情况
        {
            try {
                st = new StringTokenizer(br.readLine());
            } catch (IOException e) {
                e.printStackTrace();
            }
        }
        return st.nextToken();
    }

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

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