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 UnshufflingaDeck {

	public static void main(String[] args) {
        FastScanner in = new FastScanner();
        PrintWriter out = new PrintWriter(System.out);
        int n = in.nextInt();
        int[] a = in.readArray(n);
        int i=0,j=0;
        boolean found = false;
        for(i=0;i<n;i++){
			for(j=i+1;j<n;j++){
				if(a[i]==a[j]+1){
					found = true; break;
				}
			}
			if(found) break;
		}
        int Q = 0;
        HashMap<Integer,ArrayList<Integer>> map = new HashMap<>();
        ArrayList<Integer> p,q,r,s;
        while(i<n-1) {
        	p=new ArrayList<>();
        	q=new ArrayList<>();
        	r=new ArrayList<>();
        	s=new ArrayList<>();
        	Q++;
        	for(int l=0;l<i;l++) p.add(a[l]);
        	int k = i+1;
        	while(a[i]<a[k]){
        		k++;
        	}
        	for(int l=i;l<k;l++) q.add(a[l]);
        	for(int l=k;l<=j;l++) r.add(a[l]);
        	for(int l=j+1;l<n;l++) s.add(a[l]);
        	int ind = 0;
        	for(int l : s) {
        		a[ind] = l;ind++;
        	}
        	for(int l : r) {
        		a[ind] = l;ind++;
        	}
        	for(int l : q) {
        		a[ind] = l;ind++;
        	}
        	for(int l : p) {
        		a[ind] = l;ind++;
        	}
        	ArrayList arr = new ArrayList<>();
        	if(p.size()!=0) arr.add(p.size());
        	if(q.size()!=0) arr.add(q.size());
        	if(r.size()!=0) arr.add(r.size());
        	if(s.size()!=0) arr.add(s.size()); 
        	map.put(Q,arr);
        	i=j=0;
        	found = false;
        	for(i=0;i<n;i++){
        		for(j=i+1;j<n;j++){
        			if(a[i]==a[j]+1){
        				found = true; break;
        			}
        		}
        		if(found) break;
        	}
        }
        out.println(Q);
        i=1;
        while(Q-->0){
        	out.print(map.get(i).size()+" ");
        	for(j=0;j<map.get(i).size();j++){
        		if(map.get(i).get(j)!=0) out.print(map.get(i).get(j)+" ");
        	}
        	out.println();i++;
        }
        out.flush();
	}
	
	static class FastScanner {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st = new StringTokenizer("");
		
		String next() {
			while(!st.hasMoreTokens())
				try { st = new StringTokenizer(br.readLine()); }
				catch(IOException e) {}
			return st.nextToken();
		}
		
		String nextLine(){
			try{ return br.readLine(); } 
			catch(IOException e) { } return "";
		}
		
		int nextInt() {
			return Integer.parseInt(next());
		}
		
		long nextLong() {
			return Long.parseLong(next());
		}
		
		int[] readArray(int n) {
			int a[] = new int[n];
			for(int i=0;i<n;i++) a[i] = nextInt();
			return a;
		}
	}

	static final Random random = new Random();

	static void ruffleSort(int[] a){
		int n = a.length;
		for(int i=0;i<n;i++){
			int j = random.nextInt(n), temp = a[j];
			a[j] = a[i]; a[i] = temp;
		}
		Arrays.sort(a);
	}
}
