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.*; import java.util.*;
public class D {
    static StringBuilder sb=new StringBuilder();
    static int ans=0;
    
    public static void main(String[] args) throws IOException {
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        int N=Integer.parseInt(br.readLine());
        
        
        StringTokenizer st=new StringTokenizer(br.readLine());
        
        int[] a=new int[N];
        int cur=1;
        for (int i = 0; i < N; i++) {
            a[i]=Integer.parseInt(st.nextToken());
            //Special Cases: n=1, etc
    //Bounds: Overflow, ?
       
        }
        while(true){
            boolean flag=false;
            x:
            for (int i = 0; i < N; i++) {
                for (int j = i+1; j < N; j++) {
                    if(a[i]==a[j]+1){
                        int s=i;
                        while(a[s+1]==a[s]+1){
                            s++;
                        }
                        flag=true; ans++;
                        if(i==0){
                            if(j==N-1){
                                sb.append(2+" "+(s+1)+" "+(N-s-1)+"\n");
                            }else{
                                sb.append(3+" "+(s+1)+" "+(j-s)+" "+(N-j-1)+"\n");
                            }
                        }else{
                            if(j==N-1){
                                sb.append(3+" "+i+" "+(s-i+1)+" "+(N-s-1)+"\n");
                            }else{
                                sb.append(4+" "+i+" "+(s-i+1)+" "+(j-s)+" "+(N-j-1)+"\n");
                            }
                        }
                        
                        int[] b=new int[N];
                        for (int k = 0; k < N-j-1; k++) {
                            b[k]=a[k+(j+1)];
                        }
                        for (int k = N-j-1; k <N-s-1; k++) {
                            b[k]=a[k-(N-j-1)+(s+1)];
                        }
                        for (int k = N-s-1; k <N-i; k++) {
                            b[k]=a[k-(N-s-1)+i];
                        }
                        for (int k = N-i; k < N; k++) {
                            b[k]=a[k-(N-i)];
                        }
                        //System.out.println(Arrays.toString(b));
                        a=b.clone(); break x;
                    }
                }
            }
            if(!flag){
                System.out.println(ans); System.out.println(sb.toString()); return;
            }
        }
    }
    
}
